Explanation:
Okay, so I need to figure out the greatest length of rope that can be saved by replacing ropes HC and HD with a single rope HP, where P is on the line between C and D. The balloon is still at H above O, and the original ropes HC and HD have lengths 150 m and 130 m respectively. The distance between C and D is 140 m.

First, let me visualize the setup. There's a point O, with A north, B west, C south, D east. So, if I imagine a coordinate system with O at the origin, then A would be along the positive y-axis, B along the negative x-axis, C along the negative y-axis, and D along the positive x-axis. The points C and D are 140 meters apart. Since they are directly south and east of O, the distance between them should be the hypotenuse of a right triangle with legs equal to the distances from O to C and O to D. Wait, but the problem says the distance between C and D is 140 m. So if I let the distance from O to C be, say, y, and from O to D be x, then the distance between C and D would be sqrt(x² + y²) = 140 m.

But maybe I need to assign coordinates to all points. Let me try that. Let me set O at (0, 0). Then, since A is directly north of O, let me assume A is at (0, a) for some a. Similarly, B is directly west of O, so B is at (-b, 0) for some b. C is directly south of O, so C is at (0, -c) for some c. D is directly east of O, so D is at (d, 0) for some d.

But the distance between C and D is 140 m. So the coordinates of C are (0, -c) and D are (d, 0). The distance between them is sqrt[(d - 0)^2 + (0 - (-c))^2] = sqrt(d² + c²) = 140. So that gives me c² + d² = 140².

Now, the balloon is at H directly above O, so H has coordinates (0, h) for some h. The ropes HA, HB, HC, HD are from H to each of the points. The lengths of HC and HD are given as 150 m and 130 m. Let me write equations for those.

HC is the distance from H(0, h) to C(0, -c). That distance is sqrt[(0 - 0)^2 + (h - (-c))^2] = sqrt[(h + c)^2] = h + c. Wait, that's interesting. Since H is above O, and C is below O, the vertical distance between H and C is h + c. So HC = h + c = 150 m. Similarly, HD is the distance from H(0, h) to D(d, 0). That distance is sqrt[(d - 0)^2 + (0 - h)^2] = sqrt(d² + h²) = 130 m.

So we have two equations:

1. h + c = 150
2. sqrt(d² + h²) = 130

Also, from the distance between C and D, we have:

3. c² + d² = 140² = 19600

So, let me note that equation 2 is sqrt(d² + h²) = 130, so squaring both sides gives d² + h² = 16900.

But from equation 3, c² + d² = 19600. If I subtract equation 2 squared from equation 3, I get:

(c² + d²) - (d² + h²) = 19600 - 16900 => c² - h² = 2700.

But from equation 1, h + c = 150. Let me write that as c = 150 - h. Then, substitute into c² - h² = 2700.

So, c² - h² = (c - h)(c + h) = (c - h)(150) = 2700. Therefore, (c - h) = 2700 / 150 = 18.

So c - h = 18. But we also have c + h = 150. So we can solve for c and h.

Adding the two equations:

(c - h) + (c + h) = 18 + 150 => 2c = 168 => c = 84. Then h = 150 - c = 150 - 84 = 66.

So c = 84 m, h = 66 m. Then, from equation 2, d² + h² = 16900. So d² = 16900 - h² = 16900 - 66². Let me compute 66²: 66*66 = 4356. Then 16900 - 4356 = 12544. Then d = sqrt(12544). Let me check sqrt(12544). Well, 112² = 12544 because 100²=10000, 110²=12100, 112²= (110 + 2)^2 = 110² + 4*110 + 4 = 12100 + 440 + 4 = 12544. So d = 112 m.

So, coordinates:

O is (0, 0)

A is (0, a). Wait, but we haven't determined a yet. Wait, but maybe we don't need a? Because the problem is about replacing HC and HD with HP where P is on CD. But maybe we need coordinates for P?

Wait, but maybe we can figure out coordinates for all points. Let me check.

We have:

- C is (0, -c) = (0, -84)

- D is (d, 0) = (112, 0)

So the line between C and D goes from (0, -84) to (112, 0). So parametric equations for CD can be written as x = 112t, y = -84 + 84t, where t ranges from 0 to 1. Wait, when t=0, x=0, y=-84 (point C); when t=1, x=112, y=0 (point D). Alternatively, maybe parameterizing differently. Let me think.

Alternatively, the line CD can be parametrized as starting at C(0, -84) and moving towards D(112, 0). So direction vector is (112 - 0, 0 - (-84)) = (112, 84). So parametric equations can be written as:

x = 0 + 112s

y = -84 + 84s

where s ∈ [0, 1]. So any point P on CD can be represented as (112s, -84 + 84s) for some s between 0 and 1.

Alternatively, since the problem says P is on the straight line between C and D, so we can parametrize it as such.

Now, the original ropes HC and HD have lengths 150 m and 130 m. The total length of these two ropes is 150 + 130 = 280 m. If we replace them with a single rope HP, then the length of HP will be less than or equal to the sum of HC and HD? Wait, no. Wait, actually, the problem says "to reduce the total length of rope used, rope HC and rope HD are to be replaced by a single rope HP". So the total length saved would be (HC + HD) - HP. Therefore, to maximize the length saved, we need to minimize HP.

Therefore, the problem reduces to finding the minimum possible length of HP where P lies on CD. Then, the maximum length saved is (150 + 130) - HP_min = 280 - HP_min.

Therefore, the key is to find the minimal HP where P is on CD.

So, H is at (0, h) = (0, 66). So coordinates of H are (0, 66).

So, we need to find the point P on CD such that the distance from H(0, 66) to P is minimized.

So, this is a classic problem of finding the shortest distance from a point to a line segment. However, since CD is a straight line, and H is a point not on CD, the minimal distance from H to CD is the perpendicular distance from H to the line CD. However, we need to check if the foot of the perpendicular lies on the segment CD. If it does, then that's the minimal distance. If not, then the minimal distance is the distance to the nearest endpoint.

Therefore, first, let me find the equation of line CD.

Points C(0, -84) and D(112, 0). Let me compute the slope of CD.

Slope m = (0 - (-84)) / (112 - 0) = 84 / 112 = 3/4.

So the equation of line CD can be written as y - (-84) = (3/4)(x - 0), so y + 84 = (3/4)x => y = (3/4)x - 84.

Alternatively, in standard form: 3x - 4y - 336 = 0. Let me check:

Starting from y = (3/4)x - 84, multiply both sides by 4: 4y = 3x - 336 => 3x - 4y - 336 = 0. Yes.

So the equation of line CD is 3x - 4y - 336 = 0.

Now, the distance from H(0, 66) to line CD is given by the formula:

Distance = |Ax + By + C| / sqrt(A² + B²), where the line is Ax + By + C = 0.

But our line is 3x - 4y - 336 = 0. So A = 3, B = -4, C = -336.

Wait, the formula is |Ax + By + C| / sqrt(A² + B²). Wait, but in standard form, it's usually written as Ax + By + C = 0. So here, 3x - 4y - 336 = 0, so A=3, B=-4, C=-336.

So plugging in H(0, 66):

Distance = |3*0 + (-4)*66 + (-336)| / sqrt(3² + (-4)^2) = |0 - 264 - 336| / 5 = |-600| / 5 = 600 / 5 = 120.

So the minimal distance from H to line CD is 120 meters. However, we need to check if the foot of the perpendicular from H to CD lies on the segment CD. If it does, then HP_min = 120 m. If not, then the minimal distance is to one of the endpoints.

So let me find the coordinates of the foot of the perpendicular from H(0, 66) to line CD.

The formula for the foot of the perpendicular from a point (x0, y0) to the line ax + by + c = 0 is:

x = x0 - a*(ax0 + by0 + c)/(a² + b²)

y = y0 - b*(ax0 + by0 + c)/(a² + b²)

Wait, but let me recall the formula. Alternatively, since we have the line CD: 3x - 4y - 336 = 0.

Alternatively, parametric equations. Let me use another method.

The line CD has slope 3/4, so the perpendicular has slope -4/3.

So the line perpendicular to CD passing through H(0, 66) is y - 66 = (-4/3)(x - 0), so y = (-4/3)x + 66.

Find intersection point between this line and CD.

CD: y = (3/4)x - 84

Perpendicular: y = (-4/3)x + 66

Set equal:

(3/4)x - 84 = (-4/3)x + 66

Multiply both sides by 12 to eliminate denominators:

12*(3/4)x - 12*84 = 12*(-4/3)x + 12*66

Which is 9x - 1008 = -16x + 792

Bring all terms to left:

9x + 16x - 1008 - 792 = 0 => 25x - 1800 = 0 => 25x = 1800 => x = 1800 / 25 = 72.

Then y = (3/4)*72 - 84 = 54 - 84 = -30.

So the foot of the perpendicular is at (72, -30).

Now, check if this point lies on segment CD.

Point C is (0, -84), D is (112, 0).

So, the x-coordinate of the foot is 72, which is between 0 and 112. The y-coordinate is -30, which is between -84 and 0. Therefore, yes, the foot lies on segment CD.

Therefore, the minimal HP is 120 meters.

Therefore, the maximum length saved is (150 + 130) - 120 = 280 - 120 = 160 meters.

Wait, but hold on. Wait, the problem says "the greatest length of rope that can be saved". So replacing HC and HD with HP, which is 120 m, so saving 150 + 130 - 120 = 160 m.

But let me check if this is correct.

But let me verify the coordinates.

So point P is (72, -30). Let me check the distance from H(0, 66) to P(72, -30).

Distance squared is (72 - 0)^2 + (-30 - 66)^2 = 72² + (-96)^2.

Compute 72²: 5184, 96²: 9216. Total is 5184 + 9216 = 14400. Square root of 14400 is 120. Yes, that's correct.

So HP is indeed 120 m. Therefore, the total length saved is 150 + 130 - 120 = 160 m.

But wait, the problem says "the greatest length of rope that can be saved". Is there a possibility that HP could be shorter? But we found the minimal HP is 120 m, so replacing HC and HD with HP would save 160 m.

But let me think again. Wait, the problem says "rope HC and rope HD are to be replaced by a single rope HP". So originally, we have two ropes: HC and HD. If we replace them with one rope HP, then the total length of rope used is HP instead of HC + HD. Therefore, the amount saved is HC + HD - HP. Therefore, to maximize the saved length, we need to minimize HP.

Since we found that the minimal HP is 120 m, then the maximum saved is 150 + 130 - 120 = 160 m.

But let me check if there's any mistake here. For example, maybe the coordinates of H? Wait, earlier we found h = 66 m. Let me check that again.

We had equations:

From HC: h + c = 150.

From HD: sqrt(d² + h²) = 130.

From CD distance: c² + d² = 140² = 19600.

Then, we found c = 84, h = 66, d = 112.

Check sqrt(d² + h²) = sqrt(112² + 66²).

Compute 112² = 12544, 66² = 4356. Sum is 12544 + 4356 = 16900. sqrt(16900) = 130. Correct.

And c² + d² = 84² + 112² = 7056 + 12544 = 19600 = 140². Correct.

So coordinates are correct.

Then, the foot of the perpendicular is at (72, -30), which is on CD. Therefore, HP is 120 m.

Therefore, the maximum length saved is 160 m.

But wait, the problem says "the greatest length of rope that can be saved". Is there a possibility that if P is not the foot of the perpendicular, but somewhere else on CD, maybe HP is shorter? But no, because the minimal distance from a point to a line is the perpendicular distance. Therefore, unless the foot is outside the segment, but in this case, it is on the segment. Therefore, 120 m is indeed the minimal HP.

Therefore, the answer is 160 meters.

But let me check again. Maybe there is a miscalculation in coordinates?

Wait, coordinates of H are (0, 66). Coordinates of P are (72, -30). So the vertical distance between H and P is 66 - (-30) = 96. The horizontal distance is 72 - 0 = 72. Then, the distance is sqrt(72² + 96²). Which is sqrt( (72^2 + 96^2) ).

But 72 and 96 are multiples of 24: 72 = 3*24, 96 = 4*24. So sqrt( (3*24)^2 + (4*24)^2 ) = 24*sqrt(9 + 16) = 24*sqrt(25) = 24*5 = 120. Correct.

Therefore, yes, HP is 120.

Therefore, the maximum length saved is 150 + 130 - 120 = 160.

But let me check if the problem says "the greatest length of rope that can be saved". So, if you replace two ropes with one, the total length saved is the original total minus the new length. So yes, 150 + 130 - HP.

But maybe there's a trick here? For example, maybe the problem is not about replacing HC and HD with HP, but replacing HC and HD with a single rope that goes from H to P, but maybe there are constraints?

Wait, the problem says: "rope HC and rope HD are to be replaced by a single rope HP where P is a point on the straight line between C and D." So, the original ropes are HA, HB, HC, HD. Now, they are replacing HC and HD with HP. So, the total length of ropes is HA + HB + HC + HD. After replacement, it's HA + HB + HP. Therefore, the saved length is (HC + HD) - HP. Therefore, yes, the maximum saved is (150 + 130) - HP_min.

Therefore, 280 - 120 = 160.

But let me check if the problem says "the greatest length of rope that can be saved". Maybe there's a different interpretation? For example, maybe they want the maximum possible saving, which would be achieved by minimizing HP. Which we did.

Alternatively, maybe there's a miscalculation in the coordinates? Let me check again.

Coordinates of C: (0, -84). Coordinates of D: (112, 0). The line CD: from (0, -84) to (112, 0). The foot of the perpendicular from H(0, 66) is (72, -30). Let me check if that point is on CD.

Parametrize CD. Let me use parameter t from 0 to 1.

x(t) = 0 + 112t

y(t) = -84 + 84t

So, when t = 0.6, x = 112 * 0.6 = 67.2, y = -84 + 84*0.6 = -84 + 50.4 = -33.6. Not 72, -30. Wait, but earlier calculation said x=72, y=-30.

Wait, maybe my parametrization is different. Let me check.

Alternatively, since the foot is at (72, -30). Let me check if this lies on CD.

The line CD can be parametrized as follows. Let me find t such that x = 72 and y = -30.

From x = 112t = 72 => t = 72 / 112 = 9 / 14 ≈ 0.642857.

Then y should be -84 + 84t = -84 + 84*(9/14) = -84 + (84/14)*9 = -84 + 6*9 = -84 + 54 = -30. Correct.

So t = 9/14, which is between 0 and 1. Therefore, yes, the foot is on CD.

Therefore, HP is indeed 120 m.

Therefore, the answer is 160 m.

But let me check if the problem says "the greatest length of rope that can be saved". Maybe there's a different way? For example, if P is not on CD but somewhere else? But the problem says P is on CD.

Alternatively, maybe the problem is in three dimensions? But the Asymptote code shows a 2D diagram, and the balloon is directly above O, so H is vertically above O. So all points are in a plane? Wait, no. Wait, the points A, B, C, D, O are on a flat field, so they are in 2D. The balloon is in the air at H directly above O, so H is in 3D space, but the ropes HA, HB, HC, HD are straight lines from H to the points on the field.

But when we calculated HC, we considered it as a straight line, which is correct. So the coordinates are correct.

Therefore, I think my answer is correct.

But let me check once more.

Original ropes HC = 150 m, HD = 130 m. If we replace them with HP, which is 120 m, then the total length saved is 150 + 130 - 120 = 160 m.

Alternatively, maybe the problem is expecting something else? For example, maybe the ropes are not straight? But no, ropes are straight.

Alternatively, maybe there's a miscalculation in the distance between C and D? Wait, the problem says the distance between C and D is 140 m. From our calculation, c = 84, d = 112, so sqrt(84² + 112²) = sqrt(7056 + 12544) = sqrt(19600) = 140. Correct.

Therefore, all steps are correct.

Therefore, the answer is 160 meters.

**Final Answer**
\boxed{160}

To determine the **greatest length of rope that can be saved**, we analyze the scenario where two ropes, $ HC $ and $ HD $, are replaced by a single rope $ HP $, with point $ P $ lying on the straight line segment $ CD $. The goal is to **minimize the length of $ HP $** to **maximize the rope saved**, since the total length of the original ropes is $ HC + HD = 150 + 130 = 280 $ meters.

---

### Step 1: Coordinate Setup

Let’s place the points in a coordinate system with $ O $ at the origin $ (0, 0) $:

- $ A $ is directly north of $ O $: $ (0, a) $
- $ B $ is directly west of $ O $: $ (-b, 0) $
- $ C $ is directly south of $ O $: $ (0, -c) $
- $ D $ is directly east of $ O $: $ (d, 0) $

Given:
- $ HC = 150 $, so the vertical distance from $ H $ to $ C $ is $ h + c = 150 $
- $ HD = 130 $, so the distance from $ H $ to $ D $ is $ \sqrt{d^2 + h^2} = 130 $
- $ CD = 140 $, so $ \sqrt{c^2 + d^2} = 140 $

From these, we solve for $ c $, $ d $, and $ h $:

1. $ h + c = 150 $ → $ c = 150 - h $
2. $ \sqrt{d^2 + h^2} = 130 $ → $ d^2 + h^2 = 16900 $
3. $ \sqrt{c^2 + d^2} = 140 $ → $ c^2 + d^2 = 19600 $

Subtracting (2) from (3):

$$
c^2 - h^2 = 2700
$$

Substitute $ c = 150 - h $ into the above:

$$
(150 - h)^2 - h^2 = 2700 \\
22500 - 300h + h^2 - h^2 = 2700 \\
22500 - 300h = 2700 \\
300h = 19800 \\
h = 66
$$

Then:
- $ c = 150 - 66 = 84 $
- $ d^2 = 16900 - 66^2 = 16900 - 4356 = 12544 \Rightarrow d = 112 $

So the coordinates are:
- $ C = (0, -84) $
- $ D = (112, 0) $
- $ H = (0, 66) $

---

### Step 2: Find the Minimum Length of $ HP $

We want to find the **shortest possible length of $ HP $** where $ P $ lies on the line segment $ CD $.

The line $ CD $ passes through $ C(0, -84) $ and $ D(112, 0) $. Its slope is:

$$
m = \frac{0 - (-84)}{112 - 0} = \frac{84}{112} = \frac{3}{4}
$$

So the equation of line $ CD $ is:

$$
y = \frac{3}{4}x - 84
$$

The perpendicular from $ H(0, 66) $ to this line has slope $ -\frac{4}{3} $, so its equation is:

$$
y - 66 = -\frac{4}{3}x \Rightarrow y = -\frac{4}{3}x + 66
$$

To find the foot of the perpendicular, solve the system:

$$
\begin{cases}
y = \frac{3}{4}x - 84 \\
y = -\frac{4}{3}x + 66
\end{cases}
$$

Set the two expressions for $ y $ equal:

$$
\frac{3}{4}x - 84 = -\frac{4}{3}x + 66
$$

Multiply through by 12 to eliminate denominators:

$$
9x - 1008 = -16x + 792 \\
25x = 1800 \Rightarrow x = 72
$$

Substitute back to find $ y $:

$$
y = \frac{3}{4}(72) - 84 = 54 - 84 = -30
$$

So the foot of the perpendicular is at $ P = (72, -30) $, which lies on segment $ CD $ (since $ x = 72 $ is between 0 and 112, and $ y = -30 $ is between -84 and 0).

---

### Step 3: Compute the Minimum $ HP $

Now compute the distance from $ H(0, 66) $ to $ P(72, -30) $:

$$
HP = \sqrt{(72 - 0)^2 + (-30 - 66)^2} = \sqrt{72^2 + (-96)^2} = \sqrt{5184 + 9216} = \sqrt{14400} = 120
$$

---

### Step 4: Compute the Maximum Length Saved

The original total length of ropes $ HC $ and $ HD $ is:

$$
HC + HD = 150 + 130 = 280
$$

After replacing them with a single rope $ HP $ of length 120, the **length saved** is:

$$
280 - 120 = \boxed{160}
$$

---

### Final Answer

$$
\boxed{160}
$$

[Assistant]
Explanation:
Okay, so I need to figure out the greatest length of rope that can be saved by replacing ropes HC and HD with a single rope HP, where P is on the line between C and D. The balloon is still at H above O, and the original ropes HC and HD have lengths 150 m and 130 m respectively. The distance between C and D is 140 m.

First, let me visualize the setup. There's a point O, with A north, B west, C south, D east. So, if I imagine a coordinate system with O at the origin, then A would be along the positive y-axis, B along the negative x-axis, C along the negative y-axis, and D along the positive x-axis. The points C and D are 140 meters apart. Since they are directly south and east of O, the distance between them should be the hypotenuse of a right triangle with legs equal to the distances from O to C and O to D. Wait, but the problem says the distance between C and D is 140 m. So if I let the distance from O to C be, say, y, and from O to D be x, then the distance between C and D would be sqrt(x² + y²) = 140 m.

But maybe I need to assign coordinates to all points. Let me try that. Let me set O at (0, 0). Then, since A is directly north of O, let me assume A is at (0, a) for some a. Similarly, B is directly west of O, so B is at (-b, 0) for some b. C is directly south of O, so C is at (0, -c) for some c. D is directly east of O, so D is at (d, 0) for some d.

But the distance between C and D is 140 m. So the coordinates of C are (0, -c) and D are (d, 0). The distance between them is sqrt[(d - 0)^2 + (0 - (-c))^2] = sqrt(d² + c²) = 140. So that gives me c² + d² = 140².

Now, the balloon is at H directly above O, so H has coordinates (0, h) for some h. The ropes HA, HB, HC, HD are from H to each of the points. The lengths of HC and HD are given as 150 m and 130 m. Let me write equations for those.

HC is the distance from H(0, h) to C(0, -c). That distance is sqrt[(0 - 0)^2 + (h - (-c))^2] = sqrt[(h + c)^2] = h + c. Wait, that's interesting. Since H is above O, and C is below O, the vertical distance between H and C is h + c. So HC = h + c = 150 m. Similarly, HD is the distance from H(0, h) to D(d, 0). That distance is sqrt[(d - 0)^2 + (0 - h)^2] = sqrt(d² + h²) = 130 m.

So we have two equations:

1. h + c = 150
2. sqrt(d² + h²) = 130

Also, from the distance between C and D, we have:

3. c² + d² = 140² = 19600

So, let me note that equation 2 is sqrt(d² + h²) = 130, so squaring both sides gives d² + h² = 16900.

But from equation 3, c² + d² = 19600. If I subtract equation 2 squared from equation 3, I get:

(c² + d²) - (d² + h²) = 19600 - 16900 => c² - h² = 2700.

But from equation 1, h + c = 150. Let me write that as c = 150 - h. Then, substitute into c² - h² = 2700.

So, c² - h² = (c - h)(c + h) = (c - h)(150) = 2700. Therefore, (c - h) = 2700 / 150 = 18.

So c - h = 18. But we also have c + h = 150. So we can solve for c and h.

Adding the two equations:

(c - h) + (c + h) = 18 + 150 => 2c = 168 => c = 84. Then h = 150 - c = 150 - 84 = 66.

So c = 84 m, h = 66 m. Then, from equation 2, d² + h² = 16900. So d² = 16900 - h² = 16900 - 66². Let me compute 66²: 66*66 = 4356. Then 16900 - 4356 = 12544. Then d = sqrt(12544). Let me check sqrt(12544). Well, 112² = 12544 because 100²=10000, 110²=12100, 112²= (110 + 2)^2 = 110² + 4*110 + 4 = 12100 + 440 + 4 = 12544. So d = 112 m.

So, coordinates:

O is (0, 0)

A is (0, a). Wait, but we haven't determined a yet. Wait, but maybe we don't need a? Because the problem is about replacing HC and HD with HP where P is on CD. But maybe we need coordinates for P?

Wait, but maybe we can figure out coordinates for all points. Let me check.

We have:

- C is (0, -c) = (0, -84)

- D is (d, 0) = (112, 0)

So the line between C and D goes from (0, -84) to (112, 0). So parametric equations for CD can be written as x = 112t, y = -84 + 84t, where t ranges from 0 to 1. Wait, when t=0, x=0, y=-84 (point C); when t=1, x=112, y=0 (point D). Alternatively, maybe parameterizing differently. Let me think.

Alternatively, the line CD can be parametrized as starting at C(0, -84) and moving towards D(112, 0). So direction vector is (112 - 0, 0 - (-84)) = (112, 84). So parametric equations can be written as:

x = 0 + 112s

y = -84 + 84s

where s ∈ [0, 1]. So any point P on CD can be represented as (112s, -84 + 84s) for some s between 0 and 1.

Alternatively, since the problem says P is on the straight line between C and D, so we can parametrize it as such.

Now, the original ropes HC and HD have lengths 150 m and 130 m. The total length of these two ropes is 150 + 130 = 280 m. If we replace them with a single rope HP, then the length of HP will be less than or equal to the sum of HC and HD? Wait, no. Wait, actually, the problem says "to reduce the total length of rope used, rope HC and rope HD are to be replaced by a single rope HP". So the total length saved would be (HC + HD) - HP. Therefore, to maximize the length saved, we need to minimize HP.

Therefore, the problem reduces to finding the minimum possible length of HP where P lies on CD. Then, the maximum length saved is (150 + 130) - HP_min = 280 - HP_min.

Therefore, the key is to find the minimal HP where P is on CD.

So, H is at (0, h) = (0, 66). So coordinates of H are (0, 66).

So, we need to find the point P on CD such that the distance from H(0, 66) to P is minimized.

So, this is a classic problem of finding the shortest distance from a point to a line segment. However, since CD is a straight line, and H is a point not on CD, the minimal distance from H to CD is the perpendicular distance from H to the line CD. However, we need to check if the foot of the perpendicular lies on the segment CD. If it does, then that's the minimal distance. If not, then the minimal distance is the distance to the nearest endpoint.

Therefore, first, let me find the equation of line CD.

Points C(0, -84) and D(112, 0). Let me compute the slope of CD.

Slope m = (0 - (-84)) / (112 - 0) = 84 / 112 = 3/4.

So the equation of line CD can be written as y - (-84) = (3/4)(x - 0), so y + 84 = (3/4)x => y = (3/4)x - 84.

Alternatively, in standard form: 3x - 4y - 336 = 0. Let me check:

Starting from y = (3/4)x - 84, multiply both sides by 4: 4y = 3x - 336 => 3x - 4y - 336 = 0. Yes.

So the equation of line CD is 3x - 4y - 336 = 0.

Now, the distance from H(0, 66) to line CD is given by the formula:

Distance = |Ax + By + C| / sqrt(A² + B²), where the line is Ax + By + C = 0.

But our line is 3x - 4y - 336 = 0. So A = 3, B = -4, C = -336.

Wait, the formula is |Ax + By + C| / sqrt(A² + B²). Wait, but in standard form, it's usually written as Ax + By + C = 0. So here, 3x - 4y - 336 = 0, so A=3, B=-4, C=-336.

So plugging in H(0, 66):

Distance = |3*0 + (-4)*66 + (-336)| / sqrt(3² + (-4)^2) = |0 - 264 - 336| / 5 = |-600| / 5 = 600 / 5 = 120.

So the minimal distance from H to line CD is 120 meters. However, we need to check if the foot of the perpendicular from H to CD lies on the segment CD. If it does, then HP_min = 120 m. If not, then the minimal distance is to one of the endpoints.

So let me find the coordinates of the foot of the perpendicular from H(0, 66) to line CD.

The formula for the foot of the perpendicular from a point (x0, y0) to the line ax + by + c = 0 is:

x = x0 - a*(ax0 + by0 + c)/(a² + b²)

y = y0 - b*(ax0 + by0 + c)/(a² + b²)

Wait, but let me recall the formula. Alternatively, since we have the line CD: 3x - 4y - 336 = 0.

Alternatively, parametric equations. Let me use another method.

The line CD has slope 3/4, so the perpendicular has slope -4/3.

So the line perpendicular to CD passing through H(0, 66) is y - 66 = (-4/3)(x - 0), so y = (-4/3)x + 66.

Find intersection point between this line and CD.

CD: y = (3/4)x - 84

Perpendicular: y = (-4/3)x + 66

Set equal:

(3/4)x - 84 = (-4/3)x + 66

Multiply both sides by 12 to eliminate denominators:

12*(3/4)x - 12*84 = 12*(-4/3)x + 12*66

Which is 9x - 1008 = -16x + 792

Bring all terms to left:

9x + 16x - 1008 - 792 = 0 => 25x - 1800 = 0 => 25x = 1800 => x = 1800 / 25 = 72.

Then y = (3/4)*72 - 84 = 54 - 84 = -30.

So the foot of the perpendicular is at (72, -30).

Now, check if this point lies on segment CD.

Point C is (0, -84), D is (112, 0).

So, the x-coordinate of the foot is 72, which is between 0 and 112. The y-coordinate is -30, which is between -84 and 0. Therefore, yes, the foot lies on segment CD.

Therefore, the minimal HP is 120 meters.

Therefore, the maximum length saved is (150 + 130) - 120 = 280 - 120 = 160 meters.

Wait, but hold on. Wait, the problem says "the greatest length of rope that can be saved". So replacing HC and HD with HP, which is 120 m, so saving 150 + 130 - 120 = 160 m.

But let me check if this is correct.

But let me verify the coordinates.

So point P is (72, -30). Let me check the distance from H(0, 66) to P(72, -30).

Distance squared is (72 - 0)^2 + (-30 - 66)^2 = 72² + (-96)^2.

Compute 72²: 5184, 96²: 9216. Total is 5184 + 9216 = 14400. Square root of 14400 is 120. Yes, that's correct.

So HP is indeed 120 m. Therefore, the total length saved is 150 + 130 - 120 = 160 m.

But wait, the problem says "the greatest length of rope that can be saved". Is there a possibility that HP could be shorter? But we found the minimal HP is 120 m, so replacing HC and HD with HP would save 160 m.

But let me think again. Wait, the problem says "rope HC and rope HD are to be replaced by a single rope HP". So originally, we have two ropes: HC and HD. If we replace them with one rope HP, then the total length of rope used is HP instead of HC + HD. Therefore, the amount saved is HC + HD - HP. Therefore, to maximize the saved length, we need to minimize HP.

Since we found that the minimal HP is 120 m, then the maximum saved is 150 + 130 - 120 = 160 m.

But let me check if there's any mistake here. For example, maybe the coordinates of H? Wait, earlier we found h = 66 m. Let me check that again.

We had equations:

From HC: h + c = 150.

From HD: sqrt(d² + h²) = 130.

From CD distance: c² + d² = 140² = 19600.

Then, we found c = 84, h = 66, d = 112.

Check sqrt(d² + h²) = sqrt(112² + 66²).

Compute 112² = 12544, 66² = 4356. Sum is 12544 + 4356 = 16900. sqrt(16900) = 130. Correct.

And c² + d² = 84² + 112² = 7056 + 12544 = 19600 = 140². Correct.

So coordinates are correct.

Then, the foot of the perpendicular is at (72, -30), which is on CD. Therefore, HP is 120 m.

Therefore, the maximum length saved is 160 m.

But wait, the problem says "the greatest length of rope that can be saved". Is there a possibility that if P is not the foot of the perpendicular, but somewhere else on CD, maybe HP is shorter? But no, because the minimal distance from a point to a line is the perpendicular distance. Therefore, unless the foot is outside the segment, but in this case, it is on the segment. Therefore, 120 m is indeed the minimal HP.

Therefore, the answer is 160 meters.

But let me check again. Maybe there is a miscalculation in coordinates?

Wait, coordinates of H are (0, 66). Coordinates of P are (72, -30). So the vertical distance between H and P is 66 - (-30) = 96. The horizontal distance is 72 - 0 = 72. Then, the distance is sqrt(72² + 96²). Which is sqrt( (72^2 + 96^2) ).

But 72 and 96 are multiples of 24: 72 = 3*24, 96 = 4*24. So sqrt( (3*24)^2 + (4*24)^2 ) = 24*sqrt(9 + 16) = 24*sqrt(25) = 24*5 = 120. Correct.

Therefore, yes, HP is 120.

Therefore, the maximum length saved is 150 + 130 - 120 = 160.

But let me check if the problem says "the greatest length of rope that can be saved". So, if you replace two ropes with one, the total length saved is the original total minus the new length. So yes, 150 + 130 - HP.

But maybe there's a trick here? For example, maybe the problem is not about replacing HC and HD with HP, but replacing HC and HD with a single rope that goes from H to P, but maybe there are constraints?

Wait, the problem says: "rope HC and rope HD are to be replaced by a single rope HP where P is a point on the straight line between C and D." So, the original ropes are HA, HB, HC, HD. Now, they are replacing HC and HD with HP. So, the total length of ropes is HA + HB + HC + HD. After replacement, it's HA + HB + HP. Therefore, the saved length is (HC + HD) - HP. Therefore, yes, the maximum saved is (150 + 130) - HP_min.

Therefore, 280 - 120 = 160.

But let me check if the problem says "the greatest length of rope that can be saved". Maybe there's a different interpretation? For example, maybe they want the maximum possible saving, which would be achieved by minimizing HP. Which we did.

Alternatively, maybe there's a miscalculation in the coordinates? Let me check again.

Coordinates of C: (0, -84). Coordinates of D: (112, 0). The line CD: from (0, -84) to (112, 0). The foot of the perpendicular from H(0, 66) is (72, -30). Let me check if that point is on CD.

Parametrize CD. Let me use parameter t from 0 to 1.

x(t) = 0 + 112t

y(t) = -84 + 84t

So, when t = 0.6, x = 112 * 0.6 = 67.2, y = -84 + 84*0.6 = -84 + 50.4 = -33.6. Not 72, -30. Wait, but earlier calculation said x=72, y=-30.

Wait, maybe my parametrization is different. Let me check.

Alternatively, since the foot is at (72, -30). Let me check if this lies on CD.

The line CD can be parametrized as follows. Let me find t such that x = 72 and y = -30.

From x = 112t = 72 => t = 72 / 112 = 9 / 14 ≈ 0.642857.

Then y should be -84 + 84t = -84 + 84*(9/14) = -84 + (84/14)*9 = -84 + 6*9 = -84 + 54 = -30. Correct.

So t = 9/14, which is between 0 and 1. Therefore, yes, the foot is on CD.

Therefore, HP is indeed 120 m.

Therefore, the answer is 160 m.

But let me check if the problem says "the greatest length of rope that can be saved". Maybe there's a different way? For example, if P is not on CD but somewhere else? But the problem says P is on CD.

Alternatively, maybe the problem is in three dimensions? But the Asymptote code shows a 2D diagram, and the balloon is directly above O, so H is vertically above O. So all points are in a plane? Wait, no. Wait, the points A, B, C, D, O are on a flat field, so they are in 2D. The balloon is in the air at H directly above O, so H is in 3D space, but the ropes HA, HB, HC, HD are straight lines from H to the points on the field.

But when we calculated HC, we considered it as a straight line, which is correct. So the coordinates are correct.

Therefore, I think my answer is correct.

But let me check once more.

Original ropes HC = 150 m, HD = 130 m. If we replace them with HP, which is 120 m, then the total length saved is 150 + 130 - 120 = 160 m.

Alternatively, maybe the problem is expecting something else? For example, maybe the ropes are not straight? But no, ropes are straight.

Alternatively, maybe there's a miscalculation in the distance between C and D? Wait, the problem says the distance between C and D is 140 m. From our calculation, c = 84, d = 112, so sqrt(84² + 112²) = sqrt(7056 + 12544) = sqrt(19600) = 140. Correct.

Therefore, all steps are correct.

Therefore, the answer is 160 meters.

**Final Answer**
\boxed{160}

To determine the **greatest length of rope that can be saved**, we analyze the scenario where two ropes, $ HC $ and $ HD $, are replaced by a single rope $ HP $, with point $ P $ lying on the straight line segment $ CD $. The goal is to **minimize the length of $ HP $** to **maximize the rope saved**, since the total length of the original ropes is $ HC + HD = 150 + 130 = 280 $ meters.

---

### Step 1: Coordinate Setup

Let’s place the points in a coordinate system with $ O $ at the origin $ (0, 0) $:

- $ A $ is directly north of $ O $: $ (0, a) $
- $ B $ is directly west of $ O $: $ (-b, 0) $
- $ C $ is directly south of $ O $: $ (0, -c) $
- $ D $ is directly east of $ O $: $ (d, 0) $

Given:
- $ HC = 150 $, so the vertical distance from $ H $ to $ C $ is $ h + c = 150 $
- $ HD = 130 $, so the distance from $ H $ to $ D $ is $ \sqrt{d^2 + h^2} = 130 $
- $ CD = 140 $, so $ \sqrt{c^2 + d^2} = 140 $

From these, we solve for $ c $, $ d $, and $ h $:

1. $ h + c = 150 $ → $ c = 150 - h $
2. $ \sqrt{d^2 + h^2} = 130 $ → $ d^2 + h^2 = 16900 $
3. $ \sqrt{c^2 + d^2} = 140 $ → $ c^2 + d^2 = 19600 $

Subtracting (2) from (3):

$$
c^2 - h^2 = 2700
$$

Substitute $ c = 150 - h $ into the above:

$$
(150 - h)^2 - h^2 = 2700 \\
22500 - 300h + h^2 - h^2 = 2700 \\
22500 - 300h = 2700 \\
300h = 19800 \\
h = 66
$$

Then:
- $ c = 150 - 66 = 84 $
- $ d^2 = 16900 - 66^2 = 16900 - 4356 = 12544 \Rightarrow d = 112 $

So the coordinates are:
- $ C = (0, -84) $
- $ D = (112, 0) $
- $ H = (0, 66) $

---

### Step 2: Find the Minimum Length of $ HP $

We want to find the **shortest possible length of $ HP $** where $ P $ lies on the line segment $ CD $.

The line $ CD $ passes through $ C(0, -84) $ and $ D(112, 0) $. Its slope is:

$$
m = \frac{0 - (-84)}{112 - 0} = \frac{84}{112} = \frac{3}{4}
$$

So the equation of line $ CD $ is:

$$
y = \frac{3}{4}x - 84
$$

The perpendicular from $ H(0, 66) $ to this line has slope $ -\frac{4}{3} $, so its equation is:

$$
y - 66 = -\frac{4}{3}x \Rightarrow y = -\frac{4}{3}x + 66
$$

To find the foot of the perpendicular, solve the system:

$$
\begin{cases}
y = \frac{3}{4}x - 84 \\
y = -\frac{4}{3}x + 66
\end{cases}
$$

Set the two expressions for $ y $ equal:

$$
\frac{3}{4}x - 84 = -\frac{4}{3}x + 66
$$

Multiply through by 12 to eliminate denominators:

$$
9x - 1008 = -16x + 792 \\
25x = 1800 \Rightarrow x = 72
$$

Substitute back to find $ y $:

$$
y = \frac{3}{4}(72) - 84 = 54 - 84 = -30
$$

So the foot of the perpendicular is at $ P = (72, -30) $, which lies on segment $ CD $ (since $ x = 72 $ is between 0 and 112, and $ y = -30 $ is between -84 and 0).

---

### Step 3: Compute the Minimum $ HP $

Now compute the distance from $ H(0, 66) $ to $ P(72, -30) $:

$$
HP = \sqrt{(72 - 0)^2 + (-30 - 66)^2} = \sqrt{72^2 + (-96)^2} = \sqrt{5184 + 9216} = \sqrt{14400} = 120
$$

---

### Step 4: Compute the Maximum Length Saved

The original total length of ropes $ HC $ and $ HD $ is:

$$
HC + HD = 150 + 130 = 280
$$

After replacing them with a single rope $ HP $ of length 120, the **length saved** is:

$$
280 - 120 = \boxed{160}
$$

---

### Final Answer

$$
\boxed{160}
$$

Solution:
To save the most rope, we must have $HP$ having minimum length.
For $HP$ to have minimum length, $HP$ must be perpendicular to $CD$. [asy]
pair C, D, H, P;
H=(90,120);
C=(0,0);
D=(140,0);
P=(90,0);
draw(H--C--D--H--P);
label("H", H, N);
label("C", C, SW);
label("D", D, SE);
label("P", P, S);
label("150", (C+H)/2, NW);
label("130", (D+H)/2, NE);
[/asy] (Among other things, we can see from this diagram that sliding $P$ away from the perpendicular position does make $HP$ longer.)
In the diagram, $HC=150$, $HD=130$ and $CD=140$.
Let $HP=x$ and $PD=a$.  Then $CP=140-a$.
By the Pythagorean Theorem in $\triangle HPC$, $x^2 + (140-a)^2 = 150^2$.
By the Pythagorean Theorem in $\triangle HPD$, $x^2+a^2 = 130^2$.
Subtracting the second equation from the first, we obtain \begin{align*}
(140-a)^2 - a^2 & = 150^2 - 130^2 \\
(19600 - 280a+a^2)-a^2 & = 5600 \\
19600 -280a & = 5600 \\
280a & = 14000 \\
a & = 50
\end{align*} Therefore, $x^2 + 90^2 = 150^2$ or $x^2 = 150^2 - 90^2 = 22500 - 8100 = 14400$ so $x =120$.
So the shortest possible rope that we can use is 120 m, which saves $130+150-120 = \boxed{160}$ m of rope.


Explanation:
Okay, so Paul and Jesse each pick a number at random from the first six primes. I need to find the probability that the sum of their numbers is even. Let me think about how to approach this.

First, let me recall the first six prime numbers. Primes are numbers greater than 1 that have no divisors other than 1 and themselves. The first six primes are 2, 3, 5, 7, 11, and 13. Let me confirm that: 2 is prime, then 3, 5, 7, 11, 13. Yep, those are the first six. So each of them has these numbers to choose from.

Now, the question is about the probability that the sum of their numbers is even. So, I need to figure out in how many cases the sum is even, divided by the total number of possible outcomes.

First, let me figure out the total number of possible outcomes. Since both Paul and Jesse are choosing a number independently, each has 6 choices. So the total number of possible pairs is 6 * 6 = 36. So the denominator of my probability will be 36.

Now, the numerator is the number of pairs where the sum is even. So I need to find how many such pairs there are.

To do this, I should recall that the sum of two numbers is even if both numbers are even or both are odd. Because even + even = even, odd + odd = even, and even + odd = odd. So, the sum is even if both numbers are even or both are odd.

Therefore, I need to find the number of pairs where both numbers are even or both are odd. So, I can calculate the number of ways both are even, plus the number of ways both are odd.

So, first, let me figure out how many even numbers are in the first six primes. The primes are 2, 3, 5, 7, 11, 13. Let me check which are even. Well, 2 is the only even prime number. All other primes are odd because any other even number would be divisible by 2. So, in the first six primes, there is only one even number: 2. The rest are odd: 3, 5, 7, 11, 13. So that's 5 odd numbers and 1 even number.

Therefore, the number of ways both Paul and Jesse choose even numbers: since there's only one even number, 2. So Paul can choose 2, and Jesse can choose 2. So that's 1 * 1 = 1 way.

Wait, but hold on. Wait, if there is only one even number, then the number of ways both choose even numbers is 1 * 1 = 1? But actually, each of them can choose any of the numbers, so if there is only one even number, then the number of ways both choose even numbers is 1 (for Paul) * 1 (for Jesse) = 1. But wait, is that correct?

Wait, no. Wait, the number of even numbers is 1. So, the number of ways Paul can choose an even number is 1, and the same for Jesse. So the number of pairs where both are even is 1 * 1 = 1. But then, the number of ways both are odd is... Well, there are 5 odd numbers. So Paul can choose any of the 5 odd numbers, and Jesse can choose any of the 5 odd numbers. So that's 5 * 5 = 25. Therefore, total number of favorable outcomes is 1 + 25 = 26. Therefore, the probability is 26/36, which simplifies to 13/18. Is that correct?

Wait, let me check again. Let me think again. So total number of possible outcomes is 6*6=36. Number of favorable outcomes: both even or both odd. Since there is only one even number, the number of ways both are even is 1*1=1. The number of ways both are odd is 5*5=25. So total favorable is 1 + 25 = 26. Therefore, probability is 26/36 = 13/18. So that would be the probability. But let me check if there's another way to think about this.

Alternatively, maybe I can think about the probability that both numbers are even or both are odd. So, the probability that both are even is (number of even numbers / total numbers)^2. Similarly for both odd. So, since there is 1 even number, probability both even is (1/6)*(1/6) = 1/36. Probability both odd is (5/6)*(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. So same result. So that seems to check out.

But let me think again. Wait, is there a possibility that I made a mistake in counting the number of even numbers? Let me check the first six primes again. First six primes: 2, 3, 5, 7, 11, 13. Yes, only 2 is even. So that's correct. So only one even number. Therefore, the rest are odd. So that seems correct.

Alternatively, maybe I can think about parity. Since the sum is even if both numbers are even or both are odd. Since there is only one even number, the chance that both are even is very low. But since there are five odd numbers, the chance that both are odd is high. So 25/36 + 1/36 = 26/36. So that seems correct.

But let me think of another approach. Let me think about all possible pairs. Let me list them. But that might take time, but maybe for verification.

First, the first six primes: 2, 3, 5, 7, 11, 13. Let me think of all possible pairs. But since there are 36 pairs, that's a lot, but maybe I can count the number of pairs where the sum is even.

Alternatively, think of the parity of each number. Since only 2 is even, the rest are odd. So, if Paul picks 2, then Jesse needs to pick 2 to get an even sum. If Paul picks an odd number, then Jesse needs to pick an odd number to get an even sum.

So, let's compute the number of favorable cases:

Case 1: Paul picks 2. Then Jesse must pick 2. There is 1 choice for Paul (2) and 1 choice for Jesse (2). So 1*1=1.

Case 2: Paul picks an odd number. There are 5 odd numbers. Then Jesse must pick an odd number. There are 5 choices for Jesse. So 5*5=25.

Total favorable cases: 1 + 25 = 26. So same as before. Therefore, probability is 26/36 = 13/18. So that seems consistent.

Alternatively, maybe I can think of the probability that the sum is even as follows: the sum is even if both numbers are even or both are odd. Since there is only one even number, the probability that both are even is (1/6)(1/6) = 1/36. The probability that both are odd is (5/6)(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. Same answer.

Alternatively, maybe I can think of the probability that the sum is even as equal to the probability that both are even plus the probability that both are odd. Which is the same as above.

Alternatively, maybe I can think of the probability that the sum is even as 1 minus the probability that the sum is odd. Let me check that. The sum is odd if one is even and the other is odd. So, the probability that the sum is odd is (probability Paul even and Jesse odd) + (probability Paul odd and Jesse even). Which is (1/6 * 5/6) + (5/6 * 1/6) = 5/36 + 5/36 = 10/36. Therefore, the probability that the sum is even is 1 - 10/36 = 26/36 = 13/18. Same answer again. So that seems to confirm.

Therefore, regardless of the approach, I get 13/18. Therefore, the probability is 13/18.

But let me check once more with actual enumeration. Let me think of the possible pairs. Since there are only 6 numbers, maybe I can think of how many even and odd numbers there are.

Number of even numbers: 1 (which is 2). Number of odd numbers: 5.

So, when Paul and Jesse pick numbers, the sum is even in two cases:

1. Both pick even numbers: only possible if both pick 2. So that's 1 case.

2. Both pick odd numbers: there are 5 odd numbers, so for Paul, 5 choices, for Jesse, 5 choices, so 5*5=25 cases.

Total cases: 1 + 25 = 26. Total possible cases: 6*6=36. So 26/36=13/18. Yep, same answer.

Alternatively, maybe I can think of the possible sums. Let me think of the numbers:

Paul's numbers: 2, 3, 5, 7, 11, 13.

Jesse's numbers: same.

So, for example, if Paul picks 2 (even), then Jesse needs to pick 2 (even) to get even sum. If Paul picks 3 (odd), Jesse needs to pick odd (3,5,7,11,13). Similarly for other odd numbers.

So, for each of Paul's choices:

- If Paul picks 2 (even): Jesse must pick 2 (even). So 1 possibility.

- If Paul picks 3 (odd): Jesse must pick odd. 5 possibilities.

- Similarly for Paul picking 5, 7, 11, 13: each time, Jesse has 5 possibilities.

So total number of favorable cases: 1 + 5 + 5 + 5 + 5 + 5 = 1 + 5*5 = 1 + 25 = 26. Yep, same as before.

Therefore, I think this is correct. Therefore, the probability is 13/18.

But let me check if there's a possibility that I made a mistake in considering the numbers. For example, maybe there are more even numbers? Wait, the first six primes are 2, 3, 5, 7, 11, 13. 2 is the only even prime. All other primes are odd. So that's correct. So there is only one even number. Therefore, the rest are odd. So that's correct.

Alternatively, maybe the problem is not about the numbers they choose, but something else? Wait, no. The problem says: "the sum of the numbers they choose is even". So, the numbers are the primes, so their sum's parity depends on their individual parities.

Therefore, I think my reasoning is correct. Therefore, the probability is 13/18.

But just to be thorough, let me think of another way. Let me think of the possible parities. Let me think of the probability that Paul's number is even or odd, same for Jesse.

Probability Paul's number is even: 1/6. Probability Paul's number is odd: 5/6.

Same for Jesse.

So, the probability that both are even: (1/6)(1/6) = 1/36.

Probability both are odd: (5/6)(5/6) = 25/36.

Therefore, total probability: 1/36 + 25/36 = 26/36 = 13/18. Same answer.

Alternatively, if I think of the sum's parity. The sum is even if both are even or both are odd. So, same thing.

Alternatively, maybe I can think of the expected value? Wait, no, the question is about probability, not expectation.

Alternatively, maybe I can think of the number of even numbers and odd numbers. Since there is 1 even and 5 odd numbers. So, the number of pairs where both are even is C(1,1)*C(1,1) = 1. Wait, but combinations? Wait, no, since they are choosing with replacement? Wait, actually, since they are choosing independently, it's not combinations, it's permutations with replacement. So, the number of ways is 1*1=1 for both even. For both odd, it's 5*5=25. So same as before.

Alternatively, if they were choosing without replacement, the calculation would be different, but the problem says "each choose a number at random from the first six primes". It doesn't specify whether they choose with or without replacement. But since they are choosing independently, I think it's with replacement. So, same number for both. So, the total number of possible outcomes is 6*6=36. So, that's correct.

Therefore, I think my answer is correct. Therefore, the probability is 13/18.

But just to be 100% sure, let me think of an example with smaller numbers. Let's say, if there was only one prime, which is 2. Then, Paul and Jesse each choose 2. The sum is 4, which is even. Probability is 1. If there were two primes: 2 and 3. Then total outcomes: 4. The pairs are (2,2): sum 4 even; (2,3): sum 5 odd; (3,2): sum 5 odd; (3,3): sum 6 even. So, 2 even sums out of 4, so probability 1/2. Let's check with our formula. Number of even numbers: 1, odd numbers: 1. So, both even: 1*1=1. Both odd: 1*1=1. Total favorable: 2. Total outcomes: 4. Probability 2/4=1/2. Correct. So that works.

Another example: first three primes: 2, 3, 5. So, 1 even, 2 odd. Total outcomes: 9. Number of favorable: both even (1*1=1) and both odd (2*2=4). Total 5. So probability 5/9. Let me check by enumerating:

Pairs:

(2,2): 4 even

(2,3): 5 odd

(2,5): 7 odd

(3,2): 5 odd

(3,3): 6 even

(3,5): 8 even

(5,2): 7 odd

(5,3): 8 even

(5,5): 10 even

So, even sums: (2,2), (3,3), (3,5), (5,3), (5,5). That's 5. So 5/9. Which matches the formula. So that works. So, in this case, the formula works. Therefore, my formula seems correct.

Therefore, going back to original problem, with first six primes, 1 even, 5 odd. So, probability is 1*1 + 5*5 over 6*6 = 26/36 = 13/18. So that seems correct.

Therefore, I think the answer is 13/18.

**Final Answer**
The probability that the sum of the numbers they choose is even is \boxed{\dfrac{13}{18}}.

To determine the probability that the sum of the numbers chosen by Paul and Jesse is even, we begin by analyzing the set of numbers they are choosing from.

---

### Step 1: Identify the First Six Prime Numbers

The first six prime numbers are:

$$
\{2, 3, 5, 7, 11, 13\}
$$

Among these, only **2** is even. The rest are odd.

So, we have:
- **1 even number**
- **5 odd numbers**

---

### Step 2: Total Number of Possible Outcomes

Since both Paul and Jesse choose a number independently from the same set of 6 numbers, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 3: Determine When the Sum is Even

The sum of two numbers is **even** if:
- **Both numbers are even**, or
- **Both numbers are odd**

Let’s compute the number of favorable outcomes for each case.

#### Case 1: Both numbers are even

There is only **1 even number** (2), so the number of ways both Paul and Jesse can choose an even number is:

$$
1 \times 1 = 1
$$

#### Case 2: Both numbers are odd

There are **5 odd numbers**, so the number of ways both can choose an odd number is:

$$
5 \times 5 = 25
$$

---

### Step 4: Total Favorable Outcomes

Adding the two favorable cases:

$$
1 + 25 = 26
$$

---

### Step 5: Compute the Probability

The probability that the sum is even is the ratio of favorable outcomes to total outcomes:

$$
\frac{26}{36} = \frac{13}{18}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{18}}
$$

[Assistant]
Explanation:
Okay, so Paul and Jesse each pick a number at random from the first six primes. I need to find the probability that the sum of their numbers is even. Let me think about how to approach this.

First, let me recall the first six prime numbers. Primes are numbers greater than 1 that have no divisors other than 1 and themselves. The first six primes are 2, 3, 5, 7, 11, and 13. Let me confirm that: 2 is prime, then 3, 5, 7, 11, 13. Yep, those are the first six. So each of them has these numbers to choose from.

Now, the question is about the probability that the sum of their numbers is even. So, I need to figure out in how many cases the sum is even, divided by the total number of possible outcomes.

First, let me figure out the total number of possible outcomes. Since both Paul and Jesse are choosing a number independently, each has 6 choices. So the total number of possible pairs is 6 * 6 = 36. So the denominator of my probability will be 36.

Now, the numerator is the number of pairs where the sum is even. So I need to find how many such pairs there are.

To do this, I should recall that the sum of two numbers is even if both numbers are even or both are odd. Because even + even = even, odd + odd = even, and even + odd = odd. So, the sum is even if both numbers are even or both are odd.

Therefore, I need to find the number of pairs where both numbers are even or both are odd. So, I can calculate the number of ways both are even, plus the number of ways both are odd.

So, first, let me figure out how many even numbers are in the first six primes. The primes are 2, 3, 5, 7, 11, 13. Let me check which are even. Well, 2 is the only even prime number. All other primes are odd because any other even number would be divisible by 2. So, in the first six primes, there is only one even number: 2. The rest are odd: 3, 5, 7, 11, 13. So that's 5 odd numbers and 1 even number.

Therefore, the number of ways both Paul and Jesse choose even numbers: since there's only one even number, 2. So Paul can choose 2, and Jesse can choose 2. So that's 1 * 1 = 1 way.

Wait, but hold on. Wait, if there is only one even number, then the number of ways both choose even numbers is 1 * 1 = 1? But actually, each of them can choose any of the numbers, so if there is only one even number, then the number of ways both choose even numbers is 1 (for Paul) * 1 (for Jesse) = 1. But wait, is that correct?

Wait, no. Wait, the number of even numbers is 1. So, the number of ways Paul can choose an even number is 1, and the same for Jesse. So the number of pairs where both are even is 1 * 1 = 1. But then, the number of ways both are odd is... Well, there are 5 odd numbers. So Paul can choose any of the 5 odd numbers, and Jesse can choose any of the 5 odd numbers. So that's 5 * 5 = 25. Therefore, total number of favorable outcomes is 1 + 25 = 26. Therefore, the probability is 26/36, which simplifies to 13/18. Is that correct?

Wait, let me check again. Let me think again. So total number of possible outcomes is 6*6=36. Number of favorable outcomes: both even or both odd. Since there is only one even number, the number of ways both are even is 1*1=1. The number of ways both are odd is 5*5=25. So total favorable is 1 + 25 = 26. Therefore, probability is 26/36 = 13/18. So that would be the probability. But let me check if there's another way to think about this.

Alternatively, maybe I can think about the probability that both numbers are even or both are odd. So, the probability that both are even is (number of even numbers / total numbers)^2. Similarly for both odd. So, since there is 1 even number, probability both even is (1/6)*(1/6) = 1/36. Probability both odd is (5/6)*(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. So same result. So that seems to check out.

But let me think again. Wait, is there a possibility that I made a mistake in counting the number of even numbers? Let me check the first six primes again. First six primes: 2, 3, 5, 7, 11, 13. Yes, only 2 is even. So that's correct. So only one even number. Therefore, the rest are odd. So that seems correct.

Alternatively, maybe I can think about parity. Since the sum is even if both numbers are even or both are odd. Since there is only one even number, the chance that both are even is very low. But since there are five odd numbers, the chance that both are odd is high. So 25/36 + 1/36 = 26/36. So that seems correct.

But let me think of another approach. Let me think about all possible pairs. Let me list them. But that might take time, but maybe for verification.

First, the first six primes: 2, 3, 5, 7, 11, 13. Let me think of all possible pairs. But since there are 36 pairs, that's a lot, but maybe I can count the number of pairs where the sum is even.

Alternatively, think of the parity of each number. Since only 2 is even, the rest are odd. So, if Paul picks 2, then Jesse needs to pick 2 to get an even sum. If Paul picks an odd number, then Jesse needs to pick an odd number to get an even sum.

So, let's compute the number of favorable cases:

Case 1: Paul picks 2. Then Jesse must pick 2. There is 1 choice for Paul (2) and 1 choice for Jesse (2). So 1*1=1.

Case 2: Paul picks an odd number. There are 5 odd numbers. Then Jesse must pick an odd number. There are 5 choices for Jesse. So 5*5=25.

Total favorable cases: 1 + 25 = 26. So same as before. Therefore, probability is 26/36 = 13/18. So that seems consistent.

Alternatively, maybe I can think of the probability that the sum is even as follows: the sum is even if both numbers are even or both are odd. Since there is only one even number, the probability that both are even is (1/6)(1/6) = 1/36. The probability that both are odd is (5/6)(5/6) = 25/36. So total probability is 1/36 + 25/36 = 26/36 = 13/18. Same answer.

Alternatively, maybe I can think of the probability that the sum is even as equal to the probability that both are even plus the probability that both are odd. Which is the same as above.

Alternatively, maybe I can think of the probability that the sum is even as 1 minus the probability that the sum is odd. Let me check that. The sum is odd if one is even and the other is odd. So, the probability that the sum is odd is (probability Paul even and Jesse odd) + (probability Paul odd and Jesse even). Which is (1/6 * 5/6) + (5/6 * 1/6) = 5/36 + 5/36 = 10/36. Therefore, the probability that the sum is even is 1 - 10/36 = 26/36 = 13/18. Same answer again. So that seems to confirm.

Therefore, regardless of the approach, I get 13/18. Therefore, the probability is 13/18.

But let me check once more with actual enumeration. Let me think of the possible pairs. Since there are only 6 numbers, maybe I can think of how many even and odd numbers there are.

Number of even numbers: 1 (which is 2). Number of odd numbers: 5.

So, when Paul and Jesse pick numbers, the sum is even in two cases:

1. Both pick even numbers: only possible if both pick 2. So that's 1 case.

2. Both pick odd numbers: there are 5 odd numbers, so for Paul, 5 choices, for Jesse, 5 choices, so 5*5=25 cases.

Total cases: 1 + 25 = 26. Total possible cases: 6*6=36. So 26/36=13/18. Yep, same answer.

Alternatively, maybe I can think of the possible sums. Let me think of the numbers:

Paul's numbers: 2, 3, 5, 7, 11, 13.

Jesse's numbers: same.

So, for example, if Paul picks 2 (even), then Jesse needs to pick 2 (even) to get even sum. If Paul picks 3 (odd), Jesse needs to pick odd (3,5,7,11,13). Similarly for other odd numbers.

So, for each of Paul's choices:

- If Paul picks 2 (even): Jesse must pick 2 (even). So 1 possibility.

- If Paul picks 3 (odd): Jesse must pick odd. 5 possibilities.

- Similarly for Paul picking 5, 7, 11, 13: each time, Jesse has 5 possibilities.

So total number of favorable cases: 1 + 5 + 5 + 5 + 5 + 5 = 1 + 5*5 = 1 + 25 = 26. Yep, same as before.

Therefore, I think this is correct. Therefore, the probability is 13/18.

But let me check if there's a possibility that I made a mistake in considering the numbers. For example, maybe there are more even numbers? Wait, the first six primes are 2, 3, 5, 7, 11, 13. 2 is the only even prime. All other primes are odd. So that's correct. So there is only one even number. Therefore, the rest are odd. So that's correct.

Alternatively, maybe the problem is not about the numbers they choose, but something else? Wait, no. The problem says: "the sum of the numbers they choose is even". So, the numbers are the primes, so their sum's parity depends on their individual parities.

Therefore, I think my reasoning is correct. Therefore, the probability is 13/18.

But just to be thorough, let me think of another way. Let me think of the possible parities. Let me think of the probability that Paul's number is even or odd, same for Jesse.

Probability Paul's number is even: 1/6. Probability Paul's number is odd: 5/6.

Same for Jesse.

So, the probability that both are even: (1/6)(1/6) = 1/36.

Probability both are odd: (5/6)(5/6) = 25/36.

Therefore, total probability: 1/36 + 25/36 = 26/36 = 13/18. Same answer.

Alternatively, if I think of the sum's parity. The sum is even if both are even or both are odd. So, same thing.

Alternatively, maybe I can think of the expected value? Wait, no, the question is about probability, not expectation.

Alternatively, maybe I can think of the number of even numbers and odd numbers. Since there is 1 even and 5 odd numbers. So, the number of pairs where both are even is C(1,1)*C(1,1) = 1. Wait, but combinations? Wait, no, since they are choosing with replacement? Wait, actually, since they are choosing independently, it's not combinations, it's permutations with replacement. So, the number of ways is 1*1=1 for both even. For both odd, it's 5*5=25. So same as before.

Alternatively, if they were choosing without replacement, the calculation would be different, but the problem says "each choose a number at random from the first six primes". It doesn't specify whether they choose with or without replacement. But since they are choosing independently, I think it's with replacement. So, same number for both. So, the total number of possible outcomes is 6*6=36. So, that's correct.

Therefore, I think my answer is correct. Therefore, the probability is 13/18.

But just to be 100% sure, let me think of an example with smaller numbers. Let's say, if there was only one prime, which is 2. Then, Paul and Jesse each choose 2. The sum is 4, which is even. Probability is 1. If there were two primes: 2 and 3. Then total outcomes: 4. The pairs are (2,2): sum 4 even; (2,3): sum 5 odd; (3,2): sum 5 odd; (3,3): sum 6 even. So, 2 even sums out of 4, so probability 1/2. Let's check with our formula. Number of even numbers: 1, odd numbers: 1. So, both even: 1*1=1. Both odd: 1*1=1. Total favorable: 2. Total outcomes: 4. Probability 2/4=1/2. Correct. So that works.

Another example: first three primes: 2, 3, 5. So, 1 even, 2 odd. Total outcomes: 9. Number of favorable: both even (1*1=1) and both odd (2*2=4). Total 5. So probability 5/9. Let me check by enumerating:

Pairs:

(2,2): 4 even

(2,3): 5 odd

(2,5): 7 odd

(3,2): 5 odd

(3,3): 6 even

(3,5): 8 even

(5,2): 7 odd

(5,3): 8 even

(5,5): 10 even

So, even sums: (2,2), (3,3), (3,5), (5,3), (5,5). That's 5. So 5/9. Which matches the formula. So that works. So, in this case, the formula works. Therefore, my formula seems correct.

Therefore, going back to original problem, with first six primes, 1 even, 5 odd. So, probability is 1*1 + 5*5 over 6*6 = 26/36 = 13/18. So that seems correct.

Therefore, I think the answer is 13/18.

**Final Answer**
The probability that the sum of the numbers they choose is even is \boxed{\dfrac{13}{18}}.

To determine the probability that the sum of the numbers chosen by Paul and Jesse is even, we begin by analyzing the set of numbers they are choosing from.

---

### Step 1: Identify the First Six Prime Numbers

The first six prime numbers are:

$$
\{2, 3, 5, 7, 11, 13\}
$$

Among these, only **2** is even. The rest are odd.

So, we have:
- **1 even number**
- **5 odd numbers**

---

### Step 2: Total Number of Possible Outcomes

Since both Paul and Jesse choose a number independently from the same set of 6 numbers, the total number of possible outcomes is:

$$
6 \times 6 = 36
$$

---

### Step 3: Determine When the Sum is Even

The sum of two numbers is **even** if:
- **Both numbers are even**, or
- **Both numbers are odd**

Let’s compute the number of favorable outcomes for each case.

#### Case 1: Both numbers are even

There is only **1 even number** (2), so the number of ways both Paul and Jesse can choose an even number is:

$$
1 \times 1 = 1
$$

#### Case 2: Both numbers are odd

There are **5 odd numbers**, so the number of ways both can choose an odd number is:

$$
5 \times 5 = 25
$$

---

### Step 4: Total Favorable Outcomes

Adding the two favorable cases:

$$
1 + 25 = 26
$$

---

### Step 5: Compute the Probability

The probability that the sum is even is the ratio of favorable outcomes to total outcomes:

$$
\frac{26}{36} = \frac{13}{18}
$$

---

### Final Answer

$$
\boxed{\dfrac{13}{18}}
$$

Solution: The only way for the sum of the numbers Paul and Jesse choose to be odd is if one of them chooses 2 and the other chooses an odd prime. There are five ways for Paul to choose 2 and Jesse to choose an odd prime, and there are five ways for Jesse to choose 2 and Paul to choose an odd prime. Since there are $6\cdot 6=36$ total possible ways for Paul and Jesse to choose their numbers, the probability that the sum of the numbers Paul and Jesse choose is NOT even is $\frac{10}{36}=\frac{5}{18}$. Therefore, the probability that the sum of the numbers Paul and Jesse choose IS even is $1-\frac{5}{18}=\boxed{\frac{13}{18}}$.


Explanation:
Okay, so I need to find the maximum value of the expression (x - y)/(x⁴ + y⁴ + 6) where x and y are real numbers. Hmm, let me think about how to approach this.

First, since the expression involves both x and y, maybe I can use calculus and find critical points by taking partial derivatives? Or maybe there's a way to simplify the problem by substitution or symmetry? Let me try to explore different methods.

Let me first consider if there's any symmetry or substitution that can make this easier. The denominator is x⁴ + y⁴ + 6, which is symmetric in x and y, but the numerator is x - y, which is antisymmetric. So maybe if I set variables such that u = x + y and v = x - y? Not sure if that helps. Alternatively, maybe set t = x/y or something like that? But since both x and y can be zero, maybe that's not the best approach.

Alternatively, maybe fix one variable and express the other in terms? For example, suppose I fix y and try to maximize with respect to x, or vice versa. But since both variables are involved, maybe calculus is the way to go.

Let me try calculus. Let me denote the function as f(x, y) = (x - y)/(x⁴ + y⁴ + 6). To find its maximum, I need to find critical points where the partial derivatives are zero.

First, compute the partial derivatives. Let me recall that for a function f(x, y), the partial derivative with respect to x is ∂f/∂x, and similarly for y.

Let me compute ∂f/∂x first. Using the quotient rule: if f = numerator / denominator, then ∂f/∂x = [num’ * den - num * den’] / den².

So, numerator is (x - y), so derivative with respect to x is 1. Denominator is x⁴ + y⁴ + 6, derivative with respect to x is 4x³. Therefore:

∂f/∂x = [1*(x⁴ + y⁴ + 6) - (x - y)*4x³] / (x⁴ + y⁴ + 6)²

Similarly, ∂f/∂y is:

Numerator derivative with respect to y is -1. Denominator derivative with respect to y is 4y³. Therefore:

∂f/∂y = [(-1)*(x⁴ + y⁴ + 6) - (x - y)*4y³] / (x⁴ + y⁴ + 6)²

To find critical points, set both partial derivatives to zero. Therefore, the numerators must be zero:

For ∂f/∂x = 0:

1*(x⁴ + y⁴ + 6) - (x - y)*4x³ = 0

Similarly, for ∂f/∂y = 0:

-1*(x⁴ + y⁴ + 6) - (x - y)*4y³ = 0

So, we have two equations:

1) x⁴ + y⁴ + 6 - 4x³(x - y) = 0

2) -x⁴ - y⁴ - 6 - 4y³(x - y) = 0

Hmm, these equations look complicated. Maybe I can add them together? Let me try:

Adding equation 1 and equation 2:

(x⁴ + y⁴ + 6 - 4x³(x - y)) + (-x⁴ - y⁴ - 6 - 4y³(x - y)) = 0 + 0

Simplify term by term:

x⁴ - x⁴ + y⁴ - y⁴ + 6 - 6 - 4x³(x - y) - 4y³(x - y) = 0

Which simplifies to:

-4(x - y)(x³ + y³) = 0

So, either x - y = 0 or x³ + y³ = 0.

Case 1: x - y = 0. Then x = y. Let's substitute x = y into the original function. Then f(x, y) becomes (x - x)/(x⁴ + x⁴ + 6) = 0/(2x⁴ + 6) = 0. So, in this case, the function is zero. Since we are looking for maximum, maybe this is not the maximum. But we need to check other cases.

Case 2: x³ + y³ = 0. Which implies that y³ = -x³, so y = -x (since cube roots are unique in real numbers). Therefore, y = -x. Let's substitute y = -x into the original function. Then f(x, y) becomes (x - (-x))/(x⁴ + (-x)^4 + 6) = (2x)/(x⁴ + x⁴ + 6) = 2x/(2x⁴ + 6) = x/(x⁴ + 3). So, now the problem reduces to maximizing x/(x⁴ + 3) over real numbers x.

So, if we can find the maximum of this single-variable function, then we can find the maximum of the original expression. Let me focus on this function: g(x) = x/(x⁴ + 3). Let me find its maximum.

To find the maximum of g(x), take derivative and set to zero. Let me compute g’(x):

Using quotient rule: [1*(x⁴ + 3) - x*(4x³)] / (x⁴ + 3)^2 = [x⁴ + 3 - 4x⁴]/(x⁴ + 3)^2 = (-3x⁴ + 3)/(x⁴ + 3)^2 = 3(1 - x⁴)/(x⁴ + 3)^2.

Set derivative equal to zero: 3(1 - x⁴) = 0 => 1 - x⁴ = 0 => x⁴ = 1 => x = ±1.

So critical points at x = 1 and x = -1. Let me check the value of g(x) at these points:

At x = 1: g(1) = 1/(1 + 3) = 1/4.

At x = -1: g(-1) = (-1)/(1 + 3) = -1/4.

Therefore, the maximum of g(x) is 1/4, and the minimum is -1/4. Therefore, in case 2, the maximum value of the original function is 1/4.

But wait, we need to check if these critical points satisfy the original equations. Wait, we derived case 2 from the critical points of the original function. So, if we found that when y = -x, then the maximum of the original function is 1/4. However, we need to check if there are other critical points from case 1, but case 1 gives zero, which is less than 1/4. Therefore, maybe 1/4 is the maximum? But let me check if there are other possibilities.

Wait, but we considered the critical points from the partial derivatives. However, maybe there are other critical points where both partial derivatives are zero? Wait, we considered the cases where either x - y = 0 or x³ + y³ = 0. But perhaps there are other solutions? Wait, when we added the two equations, we found that either x - y = 0 or x³ + y³ = 0. Therefore, all critical points must satisfy one of these two cases. Therefore, the only critical points are either x = y (which gives zero) or y = -x (which gives maximum 1/4). Therefore, is 1/4 the maximum?

But wait, let me check if there are other possibilities. For example, maybe when both equations are satisfied? For example, if x - y = 0 and x³ + y³ = 0. If x = y, then x³ + y³ = 2x³ = 0 => x = 0. Therefore, the only point common to both cases is (0, 0). But at (0, 0), the original function is 0/(0 + 0 + 6) = 0, which is already considered in case 1.

Therefore, the maximum value is 1/4? Wait, but let me check if there are other critical points. For example, suppose that the equations from the partial derivatives are satisfied without x - y = 0 or x³ + y³ = 0? But according to the previous step, adding the two equations gives -4(x - y)(x³ + y³) = 0, so either x - y = 0 or x³ + y³ = 0. Therefore, there are no other critical points. Therefore, the maximum is 1/4?

But let me check with some examples. For example, take x = 1, y = -1. Then the original expression is (1 - (-1))/(1⁴ + (-1)^4 + 6) = 2/(1 + 1 + 6) = 2/8 = 1/4. So that's correct. What if I take x = 2, y = -2? Then (2 - (-2))/(16 + 16 + 6) = 4/38 ≈ 0.105, which is less than 1/4. If I take x = 1/2, y = -1/2: (1/2 - (-1/2))/( (1/16) + (1/16) + 6 ) = 1 / (1/8 + 6) = 1 / (49/8) = 8/49 ≈ 0.163, which is still less than 1/4. So seems like 1/4 is the maximum.

But wait, maybe there are other critical points where the partial derivatives are zero? For example, suppose that x and y are not related by y = -x or x = y. But according to the previous analysis, adding the two equations gives that either x - y = 0 or x³ + y³ = 0. Therefore, there are no other critical points. Therefore, the maximum is 1/4.

But let me check another approach. Maybe using substitution. Let me suppose that we set t = x - y, and try to maximize t / (x⁴ + y⁴ + 6). But how to relate t with x and y? Maybe not straightforward. Alternatively, maybe use the method of Lagrange multipliers? Let me try that.

Let me consider maximizing f(x, y) = (x - y) subject to the constraint g(x, y) = x⁴ + y⁴ + 6 = k, but since the denominator is always positive, maybe not helpful. Alternatively, since the function is (x - y)/D, where D is positive, maybe we can think of maximizing x - y while minimizing D? But it's a ratio, so it's not straightforward.

Alternatively, maybe use substitution. Let me assume that y = -x + t, but not sure. Wait, earlier approach seems to have given 1/4 as the maximum. But let me check if there are other possibilities. For example, suppose that x and y are not related by y = -x. Let me try some numbers. Suppose x is large, then x - y is... if y is fixed, then as x approaches infinity, the numerator is x - y ~ x, denominator is x⁴ + ... which tends to infinity, so the whole expression tends to zero. Similarly, if y is large, then if x is fixed, the numerator is x - y ~ -y, denominator is y⁴ + ... which tends to infinity, so the expression tends to zero. Therefore, the maximum is likely attained at some finite point.

Alternatively, maybe consider symmetry. Let me think about substituting variables. Let me set u = x + y and v = x - y. Then, x = (u + v)/2, y = (u - v)/2. Then, x⁴ + y⁴ can be expressed in terms of u and v. But this might get complicated, but let me try.

First, compute x⁴ + y⁴. Let me compute x⁴ + y⁴. Let me note that x⁴ + y⁴ = (x²)^2 + (y²)^2. Alternatively, express in terms of u and v. Let me compute x² + y² and x² - y².

x² + y² = [(u + v)/2]^2 + [(u - v)/2]^2 = [ (u² + 2uv + v²) + (u² - 2uv + v²) ] / 4 = [2u² + 2v²]/4 = (u² + v²)/2.

Similarly, x² - y² = [(u + v)/2]^2 - [(u - v)/2]^2 = [ (u² + 2uv + v²) - (u² - 2uv + v²) ] / 4 = [4uv]/4 = uv.

But x⁴ + y⁴ can be written as (x²)^2 + (y²)^2 = (x² + y²)^2 - 2x²y².

So, x⁴ + y⁴ = [(u² + v²)/2]^2 - 2x²y².

But x²y² = [(u + v)/2 * (u - v)/2]^2 = [(u² - v²)/4]^2 = (u² - v²)^2 / 16.

Therefore, x⁴ + y⁴ = [(u² + v²)^2]/4 - 2*(u² - v²)^2 / 16 = [(u² + v²)^2]/4 - (u² - v²)^2 / 8.

Let me compute this:

Let me write both terms with denominator 8:

= 2(u² + v²)^2 / 8 - (u² - v²)^2 / 8

= [2(u² + v²)^2 - (u² - v²)^2] / 8

Expanding numerator:

First, expand 2(u² + v²)^2:

2(u⁴ + 2u²v² + v⁴) = 2u⁴ + 4u²v² + 2v⁴

Then subtract (u² - v²)^2 = u⁴ - 2u²v² + v⁴:

So total numerator is 2u⁴ + 4u²v² + 2v⁴ - u⁴ + 2u²v² - v⁴ = (2u⁴ - u⁴) + (4u²v² + 2u²v²) + (2v⁴ - v⁴) = u⁴ + 6u²v² + v⁴.

Therefore, x⁴ + y⁴ = [u⁴ + 6u²v² + v⁴]/8.

Therefore, the original expression becomes:

(x - y)/[x⁴ + y⁴ + 6] = v / [ (u⁴ + 6u²v² + v⁴)/8 + 6 ] = 8v / [u⁴ + 6u²v² + v⁴ + 48].

Hmm, not sure if this helps. Maybe not. Let me think. Since we have u and v, but the expression is in terms of u and v. However, since we are trying to maximize v / [something], maybe we can consider fixing u or something? Not sure. Maybe this approach is more complicated. Let me go back.

Earlier, we found that the maximum is 1/4. But let me check if there are other critical points. For example, suppose that x and y are not related by y = -x. Wait, according to the previous analysis, all critical points must satisfy either x = y or y = -x. Therefore, there are no other critical points. Therefore, the maximum is 1/4.

But let me check another way. Suppose I set y = -x + t, but maybe not helpful. Alternatively, think of the original function. Let me consider that for maximum value, the numerator x - y is positive, so we can assume that x > y. Let me suppose that x > y. Then, maybe set variables such that x = y + t, where t > 0. Then, substitute into the function. Let me try that.

Let me set x = y + t, where t > 0. Then, the function becomes:

[(y + t) - y]/[(y + t)^4 + y^4 + 6] = t / [ (y + t)^4 + y^4 + 6 ]

Now, we need to maximize this over real numbers y and t > 0. Maybe fix t and find optimal y? Or fix y and find optimal t? Not sure. Alternatively, for fixed t, find y that minimizes the denominator. But since the denominator is (y + t)^4 + y^4 + 6, which is minimized when y is chosen such that derivative with respect to y is zero. Let me compute derivative of denominator with respect to y:

d/dy [ (y + t)^4 + y^4 + 6 ] = 4(y + t)^3 + 4y^3.

Set to zero: 4(y + t)^3 + 4y^3 = 0 => (y + t)^3 + y^3 = 0 => (y + t)^3 = -y^3 => y + t = -y => t = -2y => y = -t/2.

Therefore, for fixed t, the minimum of the denominator occurs at y = -t/2. Therefore, substituting back, the minimal denominator is:

[ (-t/2 + t)^4 + (-t/2)^4 + 6 ] = [ (t/2)^4 + (t/2)^4 + 6 ] = 2*(t^4)/16 + 6 = t^4 / 8 + 6.

Therefore, the maximum value for fixed t is t / (t^4 / 8 + 6) = 8t / (t^4 + 48). Therefore, now the problem reduces to maximizing 8t / (t^4 + 48) over t > 0.

So, let me define h(t) = 8t / (t^4 + 48). Find its maximum for t > 0. Take derivative:

h’(t) = [8(t^4 + 48) - 8t*(4t^3)] / (t^4 + 48)^2 = [8t^4 + 384 - 32t^4] / (t^4 + 48)^2 = [ -24t^4 + 384 ] / (t^4 + 48)^2 = -24(t^4 - 16)/ (t^4 + 48)^2.

Set derivative equal to zero: -24(t^4 - 16) = 0 => t^4 = 16 => t = 2 (since t > 0). Therefore, maximum occurs at t = 2. Then, h(2) = 8*2 / (16 + 48) = 16 / 64 = 1/4.

So, same result as before. Therefore, the maximum value is 1/4. Therefore, this confirms the previous result. Therefore, the maximum value is 1/4.

But let me check if there are other possibilities. For example, if we consider t negative? But since we assumed t > 0 (because x > y), but if t is negative, then x < y, and the numerator would be negative, so the maximum would be achieved at positive t. Therefore, the maximum is indeed 1/4.

Alternatively, let me check if there are other critical points when considering different substitutions. But since both methods lead to 1/4, and the critical points analysis also leads to 1/4, I think that is the correct answer.

But just to be thorough, let me check another approach. Let me consider using the AM-GM inequality or other inequalities.

The denominator is x⁴ + y⁴ + 6. Let me note that x⁴ + y⁴ ≥ 2x²y² by AM-GM. Therefore, denominator ≥ 2x²y² + 6. But not sure if helpful.

Alternatively, since we have x - y in the numerator, maybe we can bound x - y in terms of x⁴ + y⁴. But not straightforward.

Alternatively, consider that for real numbers, x⁴ ≥ 0 and y⁴ ≥ 0, so denominator is always ≥ 6. Therefore, the maximum value of the expression is at most (x - y)/6. But since x - y can be as large as... Well, if x is large and y is small, but denominator also becomes large. So not helpful.

Alternatively, think of the function as (x - y)/D, where D = x⁴ + y⁴ + 6. Suppose we fix x - y = k, then try to minimize D. Then, the maximum value of k/D would be k divided by the minimal D for that k. But how to relate k and D?

Alternatively, for fixed k = x - y, we can write y = x - k. Then, substitute into D: x⁴ + (x - k)^4 + 6. Then, we need to minimize this expression over x. Then, the maximum value of k / [min D]. But this seems similar to the previous substitution approach.

But since we already did that and found 1/4, maybe this is redundant.

Alternatively, think of the original problem as an optimization problem with variables x and y. Since we have found through multiple methods that the maximum is 1/4, and verified with examples, maybe that is the answer.

But just to be 100% sure, let me check if there are any other critical points. For example, suppose that in case 2, when y = -x, we found maximum 1/4. But suppose there are other points where the partial derivatives are zero. Wait, according to the previous equations, when we added the two partial derivatives, we found that either x = y or y = -x. Therefore, all critical points are covered. Therefore, there are no other critical points. Therefore, the maximum is indeed 1/4.

Therefore, I think the answer is 1/4.

**Final Answer**
\boxed{\dfrac{1}{4}}

To find the maximum value of the expression

$$
\frac{x - y}{x^4 + y^4 + 6}
$$

over all real numbers $ x $ and $ y $, we proceed by analyzing the structure of the function and applying calculus techniques.

---

### Step 1: Understanding the Function

Let us define the function:

$$
f(x, y) = \frac{x - y}{x^4 + y^4 + 6}
$$

The numerator is linear in $ x $ and $ y $, while the denominator is a sum of even powers of $ x $ and $ y $, plus a constant. Since the denominator is always positive (as $ x^4 + y^4 \geq 0 $), the function is well-defined for all real $ x, y $.

---

### Step 2: Critical Points via Partial Derivatives

To find the maximum, we compute the partial derivatives of $ f(x, y) $ and set them to zero.

Let’s denote the numerator as $ N = x - y $ and the denominator as $ D = x^4 + y^4 + 6 $. Then:

$$
f(x, y) = \frac{N}{D}
$$

Using the quotient rule, the partial derivatives are:

$$
\frac{\partial f}{\partial x} = \frac{D \cdot \frac{\partial N}{\partial x} - N \cdot \frac{\partial D}{\partial x}}{D^2} = \frac{D - (x - y)(4x^3)}{D^2}
$$

$$
\frac{\partial f}{\partial y} = \frac{D \cdot \frac{\partial N}{\partial y} - N \cdot \frac{\partial D}{\partial y}}{D^2} = \frac{-D - (x - y)(4y^3)}{D^2}
$$

Setting both partial derivatives to zero gives:

$$
D - 4x^3(x - y) = 0 \quad \text{(1)}
$$
$$
-D - 4y^3(x - y) = 0 \quad \text{(2)}
$$

Adding equations (1) and (2):

$$
-4(x - y)(x^3 + y^3) = 0
$$

This implies either:

- $ x - y = 0 $, or
- $ x^3 + y^3 = 0 $

---

### Step 3: Case Analysis

#### **Case 1: $ x = y $**

Then $ f(x, y) = \frac{x - x}{x^4 + x^4 + 6} = 0 $, so the function is zero in this case.

#### **Case 2: $ x^3 + y^3 = 0 \Rightarrow y = -x $**

Substitute $ y = -x $ into the function:

$$
f(x, -x) = \frac{x - (-x)}{x^4 + (-x)^4 + 6} = \frac{2x}{2x^4 + 6} = \frac{x}{x^4 + 3}
$$

Now define:

$$
g(x) = \frac{x}{x^4 + 3}
$$

We now find the maximum of $ g(x) $ over all real $ x $.

---

### Step 4: Maximizing $ g(x) = \frac{x}{x^4 + 3} $

Take the derivative:

$$
g'(x) = \frac{(x^4 + 3) - x(4x^3)}{(x^4 + 3)^2} = \frac{x^4 + 3 - 4x^4}{(x^4 + 3)^2} = \frac{-3x^4 + 3}{(x^4 + 3)^2}
$$

Set $ g'(x) = 0 $:

$$
-3x^4 + 3 = 0 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1
$$

Evaluate $ g(x) $ at these points:

- $ g(1) = \frac{1}{1 + 3} = \frac{1}{4} $
- $ g(-1) = \frac{-1}{1 + 3} = -\frac{1}{4} $

So the maximum value of $ g(x) $ is $ \frac{1}{4} $, and the minimum is $ -\frac{1}{4} $.

---

### Step 5: Conclusion

From the analysis, the only critical points that yield non-zero values are when $ y = -x $, and the maximum value of the original function is:

$$
\boxed{\dfrac{1}{4}}
$$

This is the maximum value of the expression over all real numbers $ x $ and $ y $.

[Assistant]
Explanation:
Okay, so I need to find the maximum value of the expression (x - y)/(x⁴ + y⁴ + 6) where x and y are real numbers. Hmm, let me think about how to approach this.

First, since the expression involves both x and y, maybe I can use calculus and find critical points by taking partial derivatives? Or maybe there's a way to simplify the problem by substitution or symmetry? Let me try to explore different methods.

Let me first consider if there's any symmetry or substitution that can make this easier. The denominator is x⁴ + y⁴ + 6, which is symmetric in x and y, but the numerator is x - y, which is antisymmetric. So maybe if I set variables such that u = x + y and v = x - y? Not sure if that helps. Alternatively, maybe set t = x/y or something like that? But since both x and y can be zero, maybe that's not the best approach.

Alternatively, maybe fix one variable and express the other in terms? For example, suppose I fix y and try to maximize with respect to x, or vice versa. But since both variables are involved, maybe calculus is the way to go.

Let me try calculus. Let me denote the function as f(x, y) = (x - y)/(x⁴ + y⁴ + 6). To find its maximum, I need to find critical points where the partial derivatives are zero.

First, compute the partial derivatives. Let me recall that for a function f(x, y), the partial derivative with respect to x is ∂f/∂x, and similarly for y.

Let me compute ∂f/∂x first. Using the quotient rule: if f = numerator / denominator, then ∂f/∂x = [num’ * den - num * den’] / den².

So, numerator is (x - y), so derivative with respect to x is 1. Denominator is x⁴ + y⁴ + 6, derivative with respect to x is 4x³. Therefore:

∂f/∂x = [1*(x⁴ + y⁴ + 6) - (x - y)*4x³] / (x⁴ + y⁴ + 6)²

Similarly, ∂f/∂y is:

Numerator derivative with respect to y is -1. Denominator derivative with respect to y is 4y³. Therefore:

∂f/∂y = [(-1)*(x⁴ + y⁴ + 6) - (x - y)*4y³] / (x⁴ + y⁴ + 6)²

To find critical points, set both partial derivatives to zero. Therefore, the numerators must be zero:

For ∂f/∂x = 0:

1*(x⁴ + y⁴ + 6) - (x - y)*4x³ = 0

Similarly, for ∂f/∂y = 0:

-1*(x⁴ + y⁴ + 6) - (x - y)*4y³ = 0

So, we have two equations:

1) x⁴ + y⁴ + 6 - 4x³(x - y) = 0

2) -x⁴ - y⁴ - 6 - 4y³(x - y) = 0

Hmm, these equations look complicated. Maybe I can add them together? Let me try:

Adding equation 1 and equation 2:

(x⁴ + y⁴ + 6 - 4x³(x - y)) + (-x⁴ - y⁴ - 6 - 4y³(x - y)) = 0 + 0

Simplify term by term:

x⁴ - x⁴ + y⁴ - y⁴ + 6 - 6 - 4x³(x - y) - 4y³(x - y) = 0

Which simplifies to:

-4(x - y)(x³ + y³) = 0

So, either x - y = 0 or x³ + y³ = 0.

Case 1: x - y = 0. Then x = y. Let's substitute x = y into the original function. Then f(x, y) becomes (x - x)/(x⁴ + x⁴ + 6) = 0/(2x⁴ + 6) = 0. So, in this case, the function is zero. Since we are looking for maximum, maybe this is not the maximum. But we need to check other cases.

Case 2: x³ + y³ = 0. Which implies that y³ = -x³, so y = -x (since cube roots are unique in real numbers). Therefore, y = -x. Let's substitute y = -x into the original function. Then f(x, y) becomes (x - (-x))/(x⁴ + (-x)^4 + 6) = (2x)/(x⁴ + x⁴ + 6) = 2x/(2x⁴ + 6) = x/(x⁴ + 3). So, now the problem reduces to maximizing x/(x⁴ + 3) over real numbers x.

So, if we can find the maximum of this single-variable function, then we can find the maximum of the original expression. Let me focus on this function: g(x) = x/(x⁴ + 3). Let me find its maximum.

To find the maximum of g(x), take derivative and set to zero. Let me compute g’(x):

Using quotient rule: [1*(x⁴ + 3) - x*(4x³)] / (x⁴ + 3)^2 = [x⁴ + 3 - 4x⁴]/(x⁴ + 3)^2 = (-3x⁴ + 3)/(x⁴ + 3)^2 = 3(1 - x⁴)/(x⁴ + 3)^2.

Set derivative equal to zero: 3(1 - x⁴) = 0 => 1 - x⁴ = 0 => x⁴ = 1 => x = ±1.

So critical points at x = 1 and x = -1. Let me check the value of g(x) at these points:

At x = 1: g(1) = 1/(1 + 3) = 1/4.

At x = -1: g(-1) = (-1)/(1 + 3) = -1/4.

Therefore, the maximum of g(x) is 1/4, and the minimum is -1/4. Therefore, in case 2, the maximum value of the original function is 1/4.

But wait, we need to check if these critical points satisfy the original equations. Wait, we derived case 2 from the critical points of the original function. So, if we found that when y = -x, then the maximum of the original function is 1/4. However, we need to check if there are other critical points from case 1, but case 1 gives zero, which is less than 1/4. Therefore, maybe 1/4 is the maximum? But let me check if there are other possibilities.

Wait, but we considered the critical points from the partial derivatives. However, maybe there are other critical points where both partial derivatives are zero? Wait, we considered the cases where either x - y = 0 or x³ + y³ = 0. But perhaps there are other solutions? Wait, when we added the two equations, we found that either x - y = 0 or x³ + y³ = 0. Therefore, all critical points must satisfy one of these two cases. Therefore, the only critical points are either x = y (which gives zero) or y = -x (which gives maximum 1/4). Therefore, is 1/4 the maximum?

But wait, let me check if there are other possibilities. For example, maybe when both equations are satisfied? For example, if x - y = 0 and x³ + y³ = 0. If x = y, then x³ + y³ = 2x³ = 0 => x = 0. Therefore, the only point common to both cases is (0, 0). But at (0, 0), the original function is 0/(0 + 0 + 6) = 0, which is already considered in case 1.

Therefore, the maximum value is 1/4? Wait, but let me check if there are other critical points. For example, suppose that the equations from the partial derivatives are satisfied without x - y = 0 or x³ + y³ = 0? But according to the previous step, adding the two equations gives -4(x - y)(x³ + y³) = 0, so either x - y = 0 or x³ + y³ = 0. Therefore, there are no other critical points. Therefore, the maximum is 1/4?

But let me check with some examples. For example, take x = 1, y = -1. Then the original expression is (1 - (-1))/(1⁴ + (-1)^4 + 6) = 2/(1 + 1 + 6) = 2/8 = 1/4. So that's correct. What if I take x = 2, y = -2? Then (2 - (-2))/(16 + 16 + 6) = 4/38 ≈ 0.105, which is less than 1/4. If I take x = 1/2, y = -1/2: (1/2 - (-1/2))/( (1/16) + (1/16) + 6 ) = 1 / (1/8 + 6) = 1 / (49/8) = 8/49 ≈ 0.163, which is still less than 1/4. So seems like 1/4 is the maximum.

But wait, maybe there are other critical points where the partial derivatives are zero? For example, suppose that x and y are not related by y = -x or x = y. But according to the previous analysis, adding the two equations gives that either x - y = 0 or x³ + y³ = 0. Therefore, there are no other critical points. Therefore, the maximum is 1/4.

But let me check another approach. Maybe using substitution. Let me suppose that we set t = x - y, and try to maximize t / (x⁴ + y⁴ + 6). But how to relate t with x and y? Maybe not straightforward. Alternatively, maybe use the method of Lagrange multipliers? Let me try that.

Let me consider maximizing f(x, y) = (x - y) subject to the constraint g(x, y) = x⁴ + y⁴ + 6 = k, but since the denominator is always positive, maybe not helpful. Alternatively, since the function is (x - y)/D, where D is positive, maybe we can think of maximizing x - y while minimizing D? But it's a ratio, so it's not straightforward.

Alternatively, maybe use substitution. Let me assume that y = -x + t, but not sure. Wait, earlier approach seems to have given 1/4 as the maximum. But let me check if there are other possibilities. For example, suppose that x and y are not related by y = -x. Let me try some numbers. Suppose x is large, then x - y is... if y is fixed, then as x approaches infinity, the numerator is x - y ~ x, denominator is x⁴ + ... which tends to infinity, so the whole expression tends to zero. Similarly, if y is large, then if x is fixed, the numerator is x - y ~ -y, denominator is y⁴ + ... which tends to infinity, so the expression tends to zero. Therefore, the maximum is likely attained at some finite point.

Alternatively, maybe consider symmetry. Let me think about substituting variables. Let me set u = x + y and v = x - y. Then, x = (u + v)/2, y = (u - v)/2. Then, x⁴ + y⁴ can be expressed in terms of u and v. But this might get complicated, but let me try.

First, compute x⁴ + y⁴. Let me compute x⁴ + y⁴. Let me note that x⁴ + y⁴ = (x²)^2 + (y²)^2. Alternatively, express in terms of u and v. Let me compute x² + y² and x² - y².

x² + y² = [(u + v)/2]^2 + [(u - v)/2]^2 = [ (u² + 2uv + v²) + (u² - 2uv + v²) ] / 4 = [2u² + 2v²]/4 = (u² + v²)/2.

Similarly, x² - y² = [(u + v)/2]^2 - [(u - v)/2]^2 = [ (u² + 2uv + v²) - (u² - 2uv + v²) ] / 4 = [4uv]/4 = uv.

But x⁴ + y⁴ can be written as (x²)^2 + (y²)^2 = (x² + y²)^2 - 2x²y².

So, x⁴ + y⁴ = [(u² + v²)/2]^2 - 2x²y².

But x²y² = [(u + v)/2 * (u - v)/2]^2 = [(u² - v²)/4]^2 = (u² - v²)^2 / 16.

Therefore, x⁴ + y⁴ = [(u² + v²)^2]/4 - 2*(u² - v²)^2 / 16 = [(u² + v²)^2]/4 - (u² - v²)^2 / 8.

Let me compute this:

Let me write both terms with denominator 8:

= 2(u² + v²)^2 / 8 - (u² - v²)^2 / 8

= [2(u² + v²)^2 - (u² - v²)^2] / 8

Expanding numerator:

First, expand 2(u² + v²)^2:

2(u⁴ + 2u²v² + v⁴) = 2u⁴ + 4u²v² + 2v⁴

Then subtract (u² - v²)^2 = u⁴ - 2u²v² + v⁴:

So total numerator is 2u⁴ + 4u²v² + 2v⁴ - u⁴ + 2u²v² - v⁴ = (2u⁴ - u⁴) + (4u²v² + 2u²v²) + (2v⁴ - v⁴) = u⁴ + 6u²v² + v⁴.

Therefore, x⁴ + y⁴ = [u⁴ + 6u²v² + v⁴]/8.

Therefore, the original expression becomes:

(x - y)/[x⁴ + y⁴ + 6] = v / [ (u⁴ + 6u²v² + v⁴)/8 + 6 ] = 8v / [u⁴ + 6u²v² + v⁴ + 48].

Hmm, not sure if this helps. Maybe not. Let me think. Since we have u and v, but the expression is in terms of u and v. However, since we are trying to maximize v / [something], maybe we can consider fixing u or something? Not sure. Maybe this approach is more complicated. Let me go back.

Earlier, we found that the maximum is 1/4. But let me check if there are other critical points. For example, suppose that x and y are not related by y = -x. Wait, according to the previous analysis, all critical points must satisfy either x = y or y = -x. Therefore, there are no other critical points. Therefore, the maximum is 1/4.

But let me check another way. Suppose I set y = -x + t, but maybe not helpful. Alternatively, think of the original function. Let me consider that for maximum value, the numerator x - y is positive, so we can assume that x > y. Let me suppose that x > y. Then, maybe set variables such that x = y + t, where t > 0. Then, substitute into the function. Let me try that.

Let me set x = y + t, where t > 0. Then, the function becomes:

[(y + t) - y]/[(y + t)^4 + y^4 + 6] = t / [ (y + t)^4 + y^4 + 6 ]

Now, we need to maximize this over real numbers y and t > 0. Maybe fix t and find optimal y? Or fix y and find optimal t? Not sure. Alternatively, for fixed t, find y that minimizes the denominator. But since the denominator is (y + t)^4 + y^4 + 6, which is minimized when y is chosen such that derivative with respect to y is zero. Let me compute derivative of denominator with respect to y:

d/dy [ (y + t)^4 + y^4 + 6 ] = 4(y + t)^3 + 4y^3.

Set to zero: 4(y + t)^3 + 4y^3 = 0 => (y + t)^3 + y^3 = 0 => (y + t)^3 = -y^3 => y + t = -y => t = -2y => y = -t/2.

Therefore, for fixed t, the minimum of the denominator occurs at y = -t/2. Therefore, substituting back, the minimal denominator is:

[ (-t/2 + t)^4 + (-t/2)^4 + 6 ] = [ (t/2)^4 + (t/2)^4 + 6 ] = 2*(t^4)/16 + 6 = t^4 / 8 + 6.

Therefore, the maximum value for fixed t is t / (t^4 / 8 + 6) = 8t / (t^4 + 48). Therefore, now the problem reduces to maximizing 8t / (t^4 + 48) over t > 0.

So, let me define h(t) = 8t / (t^4 + 48). Find its maximum for t > 0. Take derivative:

h’(t) = [8(t^4 + 48) - 8t*(4t^3)] / (t^4 + 48)^2 = [8t^4 + 384 - 32t^4] / (t^4 + 48)^2 = [ -24t^4 + 384 ] / (t^4 + 48)^2 = -24(t^4 - 16)/ (t^4 + 48)^2.

Set derivative equal to zero: -24(t^4 - 16) = 0 => t^4 = 16 => t = 2 (since t > 0). Therefore, maximum occurs at t = 2. Then, h(2) = 8*2 / (16 + 48) = 16 / 64 = 1/4.

So, same result as before. Therefore, the maximum value is 1/4. Therefore, this confirms the previous result. Therefore, the maximum value is 1/4.

But let me check if there are other possibilities. For example, if we consider t negative? But since we assumed t > 0 (because x > y), but if t is negative, then x < y, and the numerator would be negative, so the maximum would be achieved at positive t. Therefore, the maximum is indeed 1/4.

Alternatively, let me check if there are other critical points when considering different substitutions. But since both methods lead to 1/4, and the critical points analysis also leads to 1/4, I think that is the correct answer.

But just to be thorough, let me check another approach. Let me consider using the AM-GM inequality or other inequalities.

The denominator is x⁴ + y⁴ + 6. Let me note that x⁴ + y⁴ ≥ 2x²y² by AM-GM. Therefore, denominator ≥ 2x²y² + 6. But not sure if helpful.

Alternatively, since we have x - y in the numerator, maybe we can bound x - y in terms of x⁴ + y⁴. But not straightforward.

Alternatively, consider that for real numbers, x⁴ ≥ 0 and y⁴ ≥ 0, so denominator is always ≥ 6. Therefore, the maximum value of the expression is at most (x - y)/6. But since x - y can be as large as... Well, if x is large and y is small, but denominator also becomes large. So not helpful.

Alternatively, think of the function as (x - y)/D, where D = x⁴ + y⁴ + 6. Suppose we fix x - y = k, then try to minimize D. Then, the maximum value of k/D would be k divided by the minimal D for that k. But how to relate k and D?

Alternatively, for fixed k = x - y, we can write y = x - k. Then, substitute into D: x⁴ + (x - k)^4 + 6. Then, we need to minimize this expression over x. Then, the maximum value of k / [min D]. But this seems similar to the previous substitution approach.

But since we already did that and found 1/4, maybe this is redundant.

Alternatively, think of the original problem as an optimization problem with variables x and y. Since we have found through multiple methods that the maximum is 1/4, and verified with examples, maybe that is the answer.

But just to be 100% sure, let me check if there are any other critical points. For example, suppose that in case 2, when y = -x, we found maximum 1/4. But suppose there are other points where the partial derivatives are zero. Wait, according to the previous equations, when we added the two partial derivatives, we found that either x = y or y = -x. Therefore, all critical points are covered. Therefore, there are no other critical points. Therefore, the maximum is indeed 1/4.

Therefore, I think the answer is 1/4.

**Final Answer**
\boxed{\dfrac{1}{4}}

To find the maximum value of the expression

$$
\frac{x - y}{x^4 + y^4 + 6}
$$

over all real numbers $ x $ and $ y $, we proceed by analyzing the structure of the function and applying calculus techniques.

---

### Step 1: Understanding the Function

Let us define the function:

$$
f(x, y) = \frac{x - y}{x^4 + y^4 + 6}
$$

The numerator is linear in $ x $ and $ y $, while the denominator is a sum of even powers of $ x $ and $ y $, plus a constant. Since the denominator is always positive (as $ x^4 + y^4 \geq 0 $), the function is well-defined for all real $ x, y $.

---

### Step 2: Critical Points via Partial Derivatives

To find the maximum, we compute the partial derivatives of $ f(x, y) $ and set them to zero.

Let’s denote the numerator as $ N = x - y $ and the denominator as $ D = x^4 + y^4 + 6 $. Then:

$$
f(x, y) = \frac{N}{D}
$$

Using the quotient rule, the partial derivatives are:

$$
\frac{\partial f}{\partial x} = \frac{D \cdot \frac{\partial N}{\partial x} - N \cdot \frac{\partial D}{\partial x}}{D^2} = \frac{D - (x - y)(4x^3)}{D^2}
$$

$$
\frac{\partial f}{\partial y} = \frac{D \cdot \frac{\partial N}{\partial y} - N \cdot \frac{\partial D}{\partial y}}{D^2} = \frac{-D - (x - y)(4y^3)}{D^2}
$$

Setting both partial derivatives to zero gives:

$$
D - 4x^3(x - y) = 0 \quad \text{(1)}
$$
$$
-D - 4y^3(x - y) = 0 \quad \text{(2)}
$$

Adding equations (1) and (2):

$$
-4(x - y)(x^3 + y^3) = 0
$$

This implies either:

- $ x - y = 0 $, or
- $ x^3 + y^3 = 0 $

---

### Step 3: Case Analysis

#### **Case 1: $ x = y $**

Then $ f(x, y) = \frac{x - x}{x^4 + x^4 + 6} = 0 $, so the function is zero in this case.

#### **Case 2: $ x^3 + y^3 = 0 \Rightarrow y = -x $**

Substitute $ y = -x $ into the function:

$$
f(x, -x) = \frac{x - (-x)}{x^4 + (-x)^4 + 6} = \frac{2x}{2x^4 + 6} = \frac{x}{x^4 + 3}
$$

Now define:

$$
g(x) = \frac{x}{x^4 + 3}
$$

We now find the maximum of $ g(x) $ over all real $ x $.

---

### Step 4: Maximizing $ g(x) = \frac{x}{x^4 + 3} $

Take the derivative:

$$
g'(x) = \frac{(x^4 + 3) - x(4x^3)}{(x^4 + 3)^2} = \frac{x^4 + 3 - 4x^4}{(x^4 + 3)^2} = \frac{-3x^4 + 3}{(x^4 + 3)^2}
$$

Set $ g'(x) = 0 $:

$$
-3x^4 + 3 = 0 \Rightarrow x^4 = 1 \Rightarrow x = \pm 1
$$

Evaluate $ g(x) $ at these points:

- $ g(1) = \frac{1}{1 + 3} = \frac{1}{4} $
- $ g(-1) = \frac{-1}{1 + 3} = -\frac{1}{4} $

So the maximum value of $ g(x) $ is $ \frac{1}{4} $, and the minimum is $ -\frac{1}{4} $.

---

### Step 5: Conclusion

From the analysis, the only critical points that yield non-zero values are when $ y = -x $, and the maximum value of the original function is:

$$
\boxed{\dfrac{1}{4}}
$$

This is the maximum value of the expression over all real numbers $ x $ and $ y $.

Solution:
Clearly, the maximum occurs when $x$ is positive and $y$ is negative.  Let $z = -y,$ so $z$ is positive, and $y = -z.$  Then
\[\frac{x - y}{x^4 + y^4 + 6} = \frac{x + z}{x^4 + z^4 + 6}.\]By AM-GM,
\[x^4 + 1 + 1 + 1 \ge 4 \sqrt[4]{x^4} = 4x,\]and
\[z^4 + 1 + 1 + 1 \ge 4 \sqrt[4]{z^4} = 4z.\]Then $x^4 + z^4 + 6 \ge 4(x + z),$ which implies
\[\frac{x + z}{x^4 + z^4 + 6} \le \frac{1}{4}.\]Equality occurs when $x = z = 1,$ so the maximum value is $\boxed{\frac{1}{4}}.$