Explanation:
Okay, so I need to find the largest possible area of a rectangular field that's enclosed in an athletic track. The track has a total length of 400 meters. The Asymptote code shows that the track consists of two straight edges of the field and two semicircles. Let me try to visualize this.

First, the field is rectangular, so it has a length and a width. The track goes around the field, meaning that the two straight parts of the track are the length of the field, and the two semicircular parts are attached to the ends of the field. Wait, actually, if it's an athletic track, usually it's like a running track with straightaways and curves. So maybe the track is made up of two straight sides and two semicircular ends. So the total length of the track would be the sum of the two straight sides and the circumference of the two semicircles.

But the problem says the track consists of two edges of the field and two semicircles. So maybe the track is the perimeter of the field? Wait, no. If the field is enclosed by the track, then the track is the boundary of the field. But the track is made up of two straight edges of the field and two semicircles. So maybe the track is like a rectangle with semicircular ends? Wait, that would make the track a full circle if you put the two semicircles together. But the track is 400 meters long.

Wait, maybe the track is like a running track where the straight parts are the length of the field, and the curved parts are the semicircles. So the total length of the track would be the sum of the two straight sides and the circumference of the two semicircles. Let me think.

Let me denote the length of the field as L and the width as W. The two straight parts of the track would each be length L, so combined they are 2L. The two semicircular parts would each have a radius equal to half the width of the field, right? Because if the width of the field is W, then the diameter of each semicircle is W, so the radius is W/2.

Therefore, the circumference of a full circle with radius W/2 would be 2π(W/2) = πW. But since there are two semicircles, that's equivalent to a full circle. Therefore, the total length of the track is 2L + πW = 400 meters.

So the problem gives me that the total length of the track is 400 meters, which is 2L + πW = 400. And I need to find the maximum area of the field, which is A = L * W.

So this is an optimization problem where I need to maximize A = L * W given the constraint 2L + πW = 400.

Okay, so I can use calculus here. Let me express one variable in terms of the other using the constraint equation. Let me solve for L in terms of W.

From 2L + πW = 400, subtract πW: 2L = 400 - πW. Then divide by 2: L = (400 - πW)/2.

So then the area A = L * W = [(400 - πW)/2] * W = (400W - πW²)/2.

So A(W) = (400W - πW²)/2.

To find the maximum area, take derivative of A with respect to W, set it equal to zero.

First, let me write A(W) as A(W) = 200W - (π/2)W².

Then dA/dW = 200 - πW.

Set derivative equal to zero: 200 - πW = 0 => πW = 200 => W = 200/π.

Then plug back into the equation for L: L = (400 - π*(200/π))/2 = (400 - 200)/2 = 200/2 = 100.

Therefore, the maximum area is A = L * W = 100 * (200/π) = 20000/π.

Wait, so is that the answer? Let me check.

Alternatively, maybe I can use the method of completing the square or use AM-GM inequality? Let me see.

But since it's a quadratic equation, calculus is straightforward. Let me verify my steps again.

Total track length: 2L + πW = 400.

Area: A = L * W.

Express L in terms of W: L = (400 - πW)/2.

Then A = [(400 - πW)/2] * W = 200W - (π/2)W².

Taking derivative: dA/dW = 200 - πW. Setting to zero gives W = 200/π.

Then L = (400 - π*(200/π))/2 = (400 - 200)/2 = 100.

So area is 100*(200/π) = 20000/π ≈ 6366.197... square meters.

But let me check if this is correct.

Alternatively, maybe I can think of this as a standard optimization problem. For a rectangle with semicircular ends, the perimeter (track length) is 2L + 2*(πr), where r is the radius of the semicircles. But since the width of the rectangle is 2r, so W = 2r. Therefore, the perimeter is 2L + 2*(πr) = 2L + π*(2r) = 2L + πW. Which matches the equation I had earlier.

So that seems correct.

Alternatively, if I use calculus, the maximum area occurs when derivative is zero, so W = 200/π.

But let me check if this is indeed a maximum. The second derivative of A with respect to W is d²A/dW² = -π, which is negative, so it is a maximum.

Therefore, the maximum area is 20000/π.

But let me check if there's another way to think about this. For example, if we consider that for a given perimeter, the maximum area is achieved by a circle. But here, the track is not a circle, but a rectangle with semicircular ends. However, maybe there's a similar principle.

Alternatively, if we think of the track as a full circle with diameter equal to the width of the rectangle. Wait, no. The track is two straight sides and two semicircles. If you put the two semicircles together, they make a full circle. So the total length of the track is 2L + circumference of a circle with diameter W.

But circumference of a circle is π*diameter, so π*W. Therefore, total length is 2L + πW = 400.

So that's the same as before.

Alternatively, if we think of the track as a circle with radius r, then the circumference would be 2πr. But in this case, the track is not a circle, but a rectangle with semicircular ends. So the total length is 2L + πW.

But if we wanted to maximize the area of the field, which is L*W.

Alternatively, maybe we can use substitution. Let me think.

Let me denote the radius of the semicircular ends as r. Then the width of the field is 2r. Then the total length of the track is 2L + 2*(πr) = 2L + 2πr = 400.

So 2L + 2πr = 400 => L + πr = 200 => L = 200 - πr.

Then the area of the field is A = L * 2r = 2r*(200 - πr) = 400r - 2πr².

Taking derivative with respect to r: dA/dr = 400 - 4πr. Setting to zero: 400 - 4πr = 0 => 4πr = 400 => r = 400/(4π) = 100/π.

Then L = 200 - π*(100/π) = 200 - 100 = 100.

Therefore, area A = 400*(100/π) - 2π*(100/π)^2. Let me compute that.

First term: 400*(100/π) = 40000/π.

Second term: 2π*(10000/π²) = 20000/π.

Therefore, A = 40000/π - 20000/π = 20000/π.

Same result as before. So that's consistent.

Therefore, the maximum area is 20000 divided by π.

But let me check if the problem says "the field" is enclosed in the track. Wait, the Asymptote code shows a rectangle with two semicircular ends? Wait, the Asymptote code draws a rectangle from (0,0) to (3,0) to (3,2) to (0,2), then draws two semicircles. Wait, the Asymptote code draws the rectangle and then two semicircles on the top and bottom? Wait, looking at the Asymptote code:

It draws the rectangle from (0,0) to (3,0) to (3,2) to (0,2). Then draws the bottom side (from (0,0) to (3,0)) and the top side (from (0,2) to (3,2)) with thick lines. Then draws an arc from (3,1) with radius 1, from -90 to 90 degrees, which would be a semicircle on the right side, going up from (3,0) to (3,2). Similarly, an arc from (0,1) with radius 1, from 90 to 270 degrees, which is a semicircle on the left side, going down from (0,2) to (0,0).

So the track is the rectangle with semicircular ends on the left and right sides? Wait, but the rectangle is from (0,0) to (3,0) to (3,2) to (0,2). So the width of the rectangle is 3 units, and the height is 2 units. Then the semicircular ends are on the left and right sides? Wait, no. The semicircles are drawn on the top and bottom? Wait, no. The arc from (3,1) with radius 1, from -90 to 90 degrees. Let me think.

An arc centered at (3,1) with radius 1, from -90 degrees to 90 degrees. Starting at angle -90 degrees, which is pointing downward, so the point would be (3,1) + (0,-1) = (3,0). Then going up to 90 degrees, which is pointing upward, so (3,1) + (0,1) = (3,2). So that's a semicircle on the right side, connecting (3,0) to (3,2). Similarly, the arc on the left side is centered at (0,1), radius 1, from 90 degrees to 270 degrees. Starting at 90 degrees, which is upward, so (0,1) + (0,1) = (0,2), then going down to 270 degrees, which is (0,1) + (0,-1) = (0,0). So that's a semicircle on the left side connecting (0,2) to (0,0).

Therefore, the track is the rectangle with semicircular ends on the left and right sides? Wait, but the rectangle is from (0,0) to (3,0) to (3,2) to (0,2). So the straight parts of the track are the top and bottom sides? Wait, but the Asymptote code draws the bottom side (from (0,0) to (3,0)) and the top side (from (0,2) to (3,2)) as thick lines. Then the semicircles on the left and right sides.

Wait, so the track is like a rectangle with semicircular ends on the left and right? But that would make the track have two straight sides (top and bottom) and two semicircular sides (left and right). But in that case, the total length of the track would be the length of the top and bottom sides (each of length 3 units in the Asymptote code) plus the circumference of the two semicircles.

But in the Asymptote code, the two semicircles on the left and right sides would each have radius 1, so their total circumference is π*2*1 = 2π. The top and bottom sides are each length 3, so total length is 2*3 + 2π.

But in the problem statement, the track consists of two edges of the field and two semicircles. So in the Asymptote code, the two edges of the field are the top and bottom sides, and the two semicircles are on the left and right. Therefore, the total length of the track is 2L + πW? Wait, in the Asymptote code, the width of the field is 2 units (from y=0 to y=2), and the length is 3 units (from x=0 to x=3).

But in that case, the two semicircles have diameter equal to the width of the field? Wait, the semicircles on the left and right sides have diameter equal to the height of the field? Wait, the semicircles on the left and right sides have radius 1, so diameter 2, which is the same as the height of the field (from y=0 to y=2). Therefore, the width of the field is 2, and the length is 3.

Therefore, in this case, the total length of the track is 2*length + π*width.

Wait, but in the Asymptote code, the track is the top, bottom, left, and right? Wait, no. The track is the bottom, top, and the two semicircles. Wait, the Asymptote code draws the bottom and top sides as thick lines, and the two semicircles on the left and right. So the track is the perimeter of the field? No, because the left and right sides are replaced by semicircles.

Therefore, the total length of the track is the length of the top and bottom sides (each of length L) plus the circumference of the two semicircles (which is π*W). Therefore, total length is 2L + πW.

Therefore, in the problem statement, the track length is 400 meters, so 2L + πW = 400.

Therefore, the area of the field is L*W, which we need to maximize.

So my previous calculations are correct.

Therefore, the maximum area is 20000/π square meters.

But let me check if that's the case.

Alternatively, maybe the problem is referring to the track as the entire perimeter of the field? But no, because the track is made up of two edges of the field and two semicircles. So the track is not the entire perimeter of the field.

Wait, if the field is enclosed in the track, then the track is the boundary of the field. But the track is made up of two edges of the field and two semicircles. So maybe the track is the perimeter of the field, but with the two straight sides and two semicircular sides.

But in that case, the perimeter of the field would be 2L + πW.

But in standard terms, the perimeter of a rectangle is 2L + 2W. But here, the track is different.

But according to the problem statement, the track consists of two edges of the field and two semicircles. So the track is not the entire perimeter of the field.

But regardless, the problem says that the length of the track is 400 meters. So we have to take that as given.

Therefore, my previous calculations are correct.

But let me check with numbers. Suppose the width W is 200/π ≈ 63.66 meters, and length L is 100 meters. Then the total track length is 2*100 + π*(200/π) = 200 + 200 = 400 meters. Correct.

Then the area is 100*(200/π) ≈ 6366.197 square meters.

Alternatively, if I take different values. For example, if W = 0, then L = 200, area is 0. If W = 400/π ≈ 127.32 meters, then L = (400 - π*(400/π))/2 = (400 - 400)/2 = 0, area is 0. So the maximum is somewhere in between.

Alternatively, if I take W = 100, then L = (400 - π*100)/2 ≈ (400 - 314.16)/2 ≈ 85.84/2 ≈ 42.92. Area is 100*42.92 ≈ 4292, which is less than 6366.

So seems like 20000/π is the maximum.

Alternatively, if I use calculus, the maximum area is when the derivative is zero, which we found.

Alternatively, maybe there's a formula for this. For a rectangle with semicircular ends, the maximum area given a fixed perimeter.

But in this case, the perimeter of the track is 2L + πW.

But if we think of it as a shape with two straight sides and two semicircular sides, then the total length is 2L + πW.

Alternatively, if we think of it as a circle, the maximum area for a given perimeter is a circle. But here, the shape is constrained to be a rectangle with semicircular ends.

But in our case, the maximum area is achieved when the rectangle is such that the length L is equal to the circumference of the semicircles? Wait, no.

Wait, in our case, when we found that W = 200/π and L = 100.

So if we think of the semicircular ends, their circumference is πW.

But if we think of the length L, which is 100.

Is there a relation between L and the semicircular parts?

Alternatively, maybe if we think of the track as a full circle with diameter equal to the width of the field. Wait, if the width is W, then the diameter of the semicircles is W, so the circumference of the full circle would be πW. But the track has two straight sides of length L and the circumference of the full circle.

But if we think of the track as a full circle with diameter W, then the circumference would be πW, but the track also has two straight sides.

But in our problem, the total length is 2L + πW = 400.

Alternatively, if we wanted to make the track a full circle, we would have 2L = 0, which is not possible.

Alternatively, maybe there's a way to think of this as a circle with radius r, but with some parts replaced by straight lines.

But perhaps not necessary.

Alternatively, maybe I can use Lagrange multipliers.

Let me try that.

We need to maximize A = L*W, subject to the constraint 2L + πW = 400.

Set up the Lagrangian: L = L*W - λ(2L + πW - 400).

Wait, actually, the Lagrangian is usually written as the function to maximize minus λ times the constraint.

So, let me define the Lagrangian as:

L = L*W - λ(2L + πW - 400)

Then take partial derivatives:

∂L/∂L = W - λ*2 = 0 => W = 2λ

∂L/∂W = L - λ*π = 0 => L = πλ

∂L/∂λ = -(2L + πW - 400) = 0 => 2L + πW = 400

From the first equation: W = 2λ

From the second equation: L = πλ

Therefore, substituting into the third equation:

2*(πλ) + π*(2λ) = 400 => 2πλ + 2πλ = 400 => 4πλ = 400 => λ = 400/(4π) = 100/π

Therefore, W = 2λ = 200/π, and L = πλ = π*(100/π) = 100.

Same result as before. Therefore, the maximum area is 100*(200/π) = 20000/π.

Therefore, regardless of the method, I get the same answer.

Therefore, I think that is the correct answer.

But let me check if there's any possible mistake in interpreting the problem.

The problem says "a rectangular field is enclosed in an athletic track". The track consists of two edges of the field and two semicircles. The length of the track is 400 meters.

Wait, maybe "length of the track" refers to the total length around the track, which would be the perimeter of the track. Which we have considered as 2L + πW.

Alternatively, maybe the length of the track refers to the length of the straightaways? But that would be ambiguous. However, given the Asymptote code, which shows the track as having two straight sides and two semicircular sides, and the problem says the length of the track is 400 meters, which is likely the total length around the track.

Therefore, I think my interpretation is correct.

Therefore, the maximum area is 20000/π square meters.

But let me check if the problem is from a standard source. Wait, 400 meters is the standard length for a running track. In reality, a standard 400m track has two straights and two semicircles. The length of the track is 400 meters.

In reality, the maximum area of the field (the inner part of the track) would be when the track is designed such that the area is maximized.

But according to our calculation, the maximum area is 20000/π ≈ 6366.197 m².

But let me check with real-world data. For example, in a standard 400m track, the radius of the semicircular parts is about 36.5 meters. Let me check.

The total length of the track is 2L + 2πr = 400.

But in reality, the length of the straightaways is usually 84.39 meters, and the radius of the semicircular parts is 36.5 meters.

Let me check: 2*84.39 + 2π*36.5 ≈ 168.78 + 2*3.1416*36.5 ≈ 168.78 + 230 ≈ 398.78 meters, which is approximately 400 meters.

So in reality, the area of the field (the inner part) would be length times width. The length is 84.39 meters, and the width is 2*radius = 73 meters. Therefore, area is 84.39*73 ≈ 6160 m².

Which is close to our calculated value of 20000/π ≈ 6366 m².

So our answer is a bit larger than the real-world example. But that's because in reality, the track has lanes, and the inner part is not the maximum area. However, in our problem, we are to find the maximum area of the field enclosed in the track, given that the total length of the track is 400 meters.

Therefore, our answer of 20000/π is correct.

Therefore, I think that is the answer.

**Final Answer**
The largest possible area of the field is \boxed{\dfrac{20000}{\pi}} square meters.

To determine the **largest possible area** of a rectangular field enclosed in an athletic track, we are given that the **total length of the track is 400 meters**. The track consists of:

- Two **straight edges** of the field (i.e., the length of the rectangle),
- Two **semicircular ends** (i.e., the width of the rectangle).

---

### Step 1: Define Variables

Let:
- $ L $ be the **length** of the field (in meters),
- $ W $ be the **width** of the field (in meters).

The **total length of the track** is the sum of:
- The two straight sides: $ 2L $,
- The two semicircular ends: since the width $ W $ is the **diameter** of the semicircles, the **total curved length** is $ \pi W $ (equivalent to the circumference of a full circle with diameter $ W $).

So, the total length of the track is:
$$
2L + \pi W = 400
$$

---

### Step 2: Express Area in Terms of One Variable

We are to **maximize the area** of the field:
$$
A = L \cdot W
$$

From the constraint $ 2L + \pi W = 400 $, solve for $ L $:
$$
L = \frac{400 - \pi W}{2}
$$

Substitute into the area formula:
$$
A = \left( \frac{400 - \pi W}{2} \right) \cdot W = \frac{400W - \pi W^2}{2}
$$

---

### Step 3: Maximize the Area

Let:
$$
A(W) = \frac{400W - \pi W^2}{2}
$$

Take the derivative with respect to $ W $:
$$
\frac{dA}{dW} = \frac{400 - 2\pi W}{2} = 200 - \pi W
$$

Set the derivative equal to zero to find the critical point:
$$
200 - \pi W = 0 \Rightarrow W = \frac{200}{\pi}
$$

Now substitute back to find $ L $:
$$
L = \frac{400 - \pi \cdot \frac{200}{\pi}}{2} = \frac{400 - 200}{2} = 100
$$

---

### Step 4: Compute the Maximum Area

$$
A = L \cdot W = 100 \cdot \frac{200}{\pi} = \frac{20000}{\pi}
$$

---

### Final Answer

$$
\boxed{\dfrac{20000}{\pi}}
$$

This is the **largest possible area** of the rectangular field enclosed in the 400-meter athletic track.

[Assistant]
Explanation:
Okay, so I need to find the largest possible area of a rectangular field that's enclosed in an athletic track. The track has a total length of 400 meters. The Asymptote code shows that the track consists of two straight edges of the field and two semicircles. Let me try to visualize this.

First, the field is rectangular, so it has a length and a width. The track goes around the field, meaning that the two straight parts of the track are the length of the field, and the two semicircular parts are attached to the ends of the field. Wait, actually, if it's an athletic track, usually it's like a running track with straightaways and curves. So maybe the track is made up of two straight sides and two semicircular ends. So the total length of the track would be the sum of the two straight sides and the circumference of the two semicircles.

But the problem says the track consists of two edges of the field and two semicircles. So maybe the track is the perimeter of the field? Wait, no. If the field is enclosed by the track, then the track is the boundary of the field. But the track is made up of two straight edges of the field and two semicircles. So maybe the track is like a rectangle with semicircular ends? Wait, that would make the track a full circle if you put the two semicircles together. But the track is 400 meters long.

Wait, maybe the track is like a running track where the straight parts are the length of the field, and the curved parts are the semicircles. So the total length of the track would be the sum of the two straight sides and the circumference of the two semicircles. Let me think.

Let me denote the length of the field as L and the width as W. The two straight parts of the track would each be length L, so combined they are 2L. The two semicircular parts would each have a radius equal to half the width of the field, right? Because if the width of the field is W, then the diameter of each semicircle is W, so the radius is W/2.

Therefore, the circumference of a full circle with radius W/2 would be 2π(W/2) = πW. But since there are two semicircles, that's equivalent to a full circle. Therefore, the total length of the track is 2L + πW = 400 meters.

So the problem gives me that the total length of the track is 400 meters, which is 2L + πW = 400. And I need to find the maximum area of the field, which is A = L * W.

So this is an optimization problem where I need to maximize A = L * W given the constraint 2L + πW = 400.

Okay, so I can use calculus here. Let me express one variable in terms of the other using the constraint equation. Let me solve for L in terms of W.

From 2L + πW = 400, subtract πW: 2L = 400 - πW. Then divide by 2: L = (400 - πW)/2.

So then the area A = L * W = [(400 - πW)/2] * W = (400W - πW²)/2.

So A(W) = (400W - πW²)/2.

To find the maximum area, take derivative of A with respect to W, set it equal to zero.

First, let me write A(W) as A(W) = 200W - (π/2)W².

Then dA/dW = 200 - πW.

Set derivative equal to zero: 200 - πW = 0 => πW = 200 => W = 200/π.

Then plug back into the equation for L: L = (400 - π*(200/π))/2 = (400 - 200)/2 = 200/2 = 100.

Therefore, the maximum area is A = L * W = 100 * (200/π) = 20000/π.

Wait, so is that the answer? Let me check.

Alternatively, maybe I can use the method of completing the square or use AM-GM inequality? Let me see.

But since it's a quadratic equation, calculus is straightforward. Let me verify my steps again.

Total track length: 2L + πW = 400.

Area: A = L * W.

Express L in terms of W: L = (400 - πW)/2.

Then A = [(400 - πW)/2] * W = 200W - (π/2)W².

Taking derivative: dA/dW = 200 - πW. Setting to zero gives W = 200/π.

Then L = (400 - π*(200/π))/2 = (400 - 200)/2 = 100.

So area is 100*(200/π) = 20000/π ≈ 6366.197... square meters.

But let me check if this is correct.

Alternatively, maybe I can think of this as a standard optimization problem. For a rectangle with semicircular ends, the perimeter (track length) is 2L + 2*(πr), where r is the radius of the semicircles. But since the width of the rectangle is 2r, so W = 2r. Therefore, the perimeter is 2L + 2*(πr) = 2L + π*(2r) = 2L + πW. Which matches the equation I had earlier.

So that seems correct.

Alternatively, if I use calculus, the maximum area occurs when derivative is zero, so W = 200/π.

But let me check if this is indeed a maximum. The second derivative of A with respect to W is d²A/dW² = -π, which is negative, so it is a maximum.

Therefore, the maximum area is 20000/π.

But let me check if there's another way to think about this. For example, if we consider that for a given perimeter, the maximum area is achieved by a circle. But here, the track is not a circle, but a rectangle with semicircular ends. However, maybe there's a similar principle.

Alternatively, if we think of the track as a full circle with diameter equal to the width of the rectangle. Wait, no. The track is two straight sides and two semicircles. If you put the two semicircles together, they make a full circle. So the total length of the track is 2L + circumference of a circle with diameter W.

But circumference of a circle is π*diameter, so π*W. Therefore, total length is 2L + πW = 400.

So that's the same as before.

Alternatively, if we think of the track as a circle with radius r, then the circumference would be 2πr. But in this case, the track is not a circle, but a rectangle with semicircular ends. So the total length is 2L + πW.

But if we wanted to maximize the area of the field, which is L*W.

Alternatively, maybe we can use substitution. Let me think.

Let me denote the radius of the semicircular ends as r. Then the width of the field is 2r. Then the total length of the track is 2L + 2*(πr) = 2L + 2πr = 400.

So 2L + 2πr = 400 => L + πr = 200 => L = 200 - πr.

Then the area of the field is A = L * 2r = 2r*(200 - πr) = 400r - 2πr².

Taking derivative with respect to r: dA/dr = 400 - 4πr. Setting to zero: 400 - 4πr = 0 => 4πr = 400 => r = 400/(4π) = 100/π.

Then L = 200 - π*(100/π) = 200 - 100 = 100.

Therefore, area A = 400*(100/π) - 2π*(100/π)^2. Let me compute that.

First term: 400*(100/π) = 40000/π.

Second term: 2π*(10000/π²) = 20000/π.

Therefore, A = 40000/π - 20000/π = 20000/π.

Same result as before. So that's consistent.

Therefore, the maximum area is 20000 divided by π.

But let me check if the problem says "the field" is enclosed in the track. Wait, the Asymptote code shows a rectangle with two semicircular ends? Wait, the Asymptote code draws a rectangle from (0,0) to (3,0) to (3,2) to (0,2), then draws two semicircles. Wait, the Asymptote code draws the rectangle and then two semicircles on the top and bottom? Wait, looking at the Asymptote code:

It draws the rectangle from (0,0) to (3,0) to (3,2) to (0,2). Then draws the bottom side (from (0,0) to (3,0)) and the top side (from (0,2) to (3,2)) with thick lines. Then draws an arc from (3,1) with radius 1, from -90 to 90 degrees, which would be a semicircle on the right side, going up from (3,0) to (3,2). Similarly, an arc from (0,1) with radius 1, from 90 to 270 degrees, which is a semicircle on the left side, going down from (0,2) to (0,0).

So the track is the rectangle with semicircular ends on the left and right sides? Wait, but the rectangle is from (0,0) to (3,0) to (3,2) to (0,2). So the width of the rectangle is 3 units, and the height is 2 units. Then the semicircular ends are on the left and right sides? Wait, no. The semicircles are drawn on the top and bottom? Wait, no. The arc from (3,1) with radius 1, from -90 to 90 degrees. Let me think.

An arc centered at (3,1) with radius 1, from -90 degrees to 90 degrees. Starting at angle -90 degrees, which is pointing downward, so the point would be (3,1) + (0,-1) = (3,0). Then going up to 90 degrees, which is pointing upward, so (3,1) + (0,1) = (3,2). So that's a semicircle on the right side, connecting (3,0) to (3,2). Similarly, the arc on the left side is centered at (0,1), radius 1, from 90 degrees to 270 degrees. Starting at 90 degrees, which is upward, so (0,1) + (0,1) = (0,2), then going down to 270 degrees, which is (0,1) + (0,-1) = (0,0). So that's a semicircle on the left side connecting (0,2) to (0,0).

Therefore, the track is the rectangle with semicircular ends on the left and right sides? Wait, but the rectangle is from (0,0) to (3,0) to (3,2) to (0,2). So the straight parts of the track are the top and bottom sides? Wait, but the Asymptote code draws the bottom side (from (0,0) to (3,0)) and the top side (from (0,2) to (3,2)) as thick lines. Then the semicircles on the left and right sides.

Wait, so the track is like a rectangle with semicircular ends on the left and right? But that would make the track have two straight sides (top and bottom) and two semicircular sides (left and right). But in that case, the total length of the track would be the length of the top and bottom sides (each of length 3 units in the Asymptote code) plus the circumference of the two semicircles.

But in the Asymptote code, the two semicircles on the left and right sides would each have radius 1, so their total circumference is π*2*1 = 2π. The top and bottom sides are each length 3, so total length is 2*3 + 2π.

But in the problem statement, the track consists of two edges of the field and two semicircles. So in the Asymptote code, the two edges of the field are the top and bottom sides, and the two semicircles are on the left and right. Therefore, the total length of the track is 2L + πW? Wait, in the Asymptote code, the width of the field is 2 units (from y=0 to y=2), and the length is 3 units (from x=0 to x=3).

But in that case, the two semicircles have diameter equal to the width of the field? Wait, the semicircles on the left and right sides have diameter equal to the height of the field? Wait, the semicircles on the left and right sides have radius 1, so diameter 2, which is the same as the height of the field (from y=0 to y=2). Therefore, the width of the field is 2, and the length is 3.

Therefore, in this case, the total length of the track is 2*length + π*width.

Wait, but in the Asymptote code, the track is the top, bottom, left, and right? Wait, no. The track is the bottom, top, and the two semicircles. Wait, the Asymptote code draws the bottom and top sides as thick lines, and the two semicircles on the left and right. So the track is the perimeter of the field? No, because the left and right sides are replaced by semicircles.

Therefore, the total length of the track is the length of the top and bottom sides (each of length L) plus the circumference of the two semicircles (which is π*W). Therefore, total length is 2L + πW.

Therefore, in the problem statement, the track length is 400 meters, so 2L + πW = 400.

Therefore, the area of the field is L*W, which we need to maximize.

So my previous calculations are correct.

Therefore, the maximum area is 20000/π square meters.

But let me check if that's the case.

Alternatively, maybe the problem is referring to the track as the entire perimeter of the field? But no, because the track is made up of two edges of the field and two semicircles. So the track is not the entire perimeter of the field.

Wait, if the field is enclosed in the track, then the track is the boundary of the field. But the track is made up of two edges of the field and two semicircles. So maybe the track is the perimeter of the field, but with the two straight sides and two semicircular sides.

But in that case, the perimeter of the field would be 2L + πW.

But in standard terms, the perimeter of a rectangle is 2L + 2W. But here, the track is different.

But according to the problem statement, the track consists of two edges of the field and two semicircles. So the track is not the entire perimeter of the field.

But regardless, the problem says that the length of the track is 400 meters. So we have to take that as given.

Therefore, my previous calculations are correct.

But let me check with numbers. Suppose the width W is 200/π ≈ 63.66 meters, and length L is 100 meters. Then the total track length is 2*100 + π*(200/π) = 200 + 200 = 400 meters. Correct.

Then the area is 100*(200/π) ≈ 6366.197 square meters.

Alternatively, if I take different values. For example, if W = 0, then L = 200, area is 0. If W = 400/π ≈ 127.32 meters, then L = (400 - π*(400/π))/2 = (400 - 400)/2 = 0, area is 0. So the maximum is somewhere in between.

Alternatively, if I take W = 100, then L = (400 - π*100)/2 ≈ (400 - 314.16)/2 ≈ 85.84/2 ≈ 42.92. Area is 100*42.92 ≈ 4292, which is less than 6366.

So seems like 20000/π is the maximum.

Alternatively, if I use calculus, the maximum area is when the derivative is zero, which we found.

Alternatively, maybe there's a formula for this. For a rectangle with semicircular ends, the maximum area given a fixed perimeter.

But in this case, the perimeter of the track is 2L + πW.

But if we think of it as a shape with two straight sides and two semicircular sides, then the total length is 2L + πW.

Alternatively, if we think of it as a circle, the maximum area for a given perimeter is a circle. But here, the shape is constrained to be a rectangle with semicircular ends.

But in our case, the maximum area is achieved when the rectangle is such that the length L is equal to the circumference of the semicircles? Wait, no.

Wait, in our case, when we found that W = 200/π and L = 100.

So if we think of the semicircular ends, their circumference is πW.

But if we think of the length L, which is 100.

Is there a relation between L and the semicircular parts?

Alternatively, maybe if we think of the track as a full circle with diameter equal to the width of the field. Wait, if the width is W, then the diameter of the semicircles is W, so the circumference of the full circle would be πW. But the track has two straight sides of length L and the circumference of the full circle.

But if we think of the track as a full circle with diameter W, then the circumference would be πW, but the track also has two straight sides.

But in our problem, the total length is 2L + πW = 400.

Alternatively, if we wanted to make the track a full circle, we would have 2L = 0, which is not possible.

Alternatively, maybe there's a way to think of this as a circle with radius r, but with some parts replaced by straight lines.

But perhaps not necessary.

Alternatively, maybe I can use Lagrange multipliers.

Let me try that.

We need to maximize A = L*W, subject to the constraint 2L + πW = 400.

Set up the Lagrangian: L = L*W - λ(2L + πW - 400).

Wait, actually, the Lagrangian is usually written as the function to maximize minus λ times the constraint.

So, let me define the Lagrangian as:

L = L*W - λ(2L + πW - 400)

Then take partial derivatives:

∂L/∂L = W - λ*2 = 0 => W = 2λ

∂L/∂W = L - λ*π = 0 => L = πλ

∂L/∂λ = -(2L + πW - 400) = 0 => 2L + πW = 400

From the first equation: W = 2λ

From the second equation: L = πλ

Therefore, substituting into the third equation:

2*(πλ) + π*(2λ) = 400 => 2πλ + 2πλ = 400 => 4πλ = 400 => λ = 400/(4π) = 100/π

Therefore, W = 2λ = 200/π, and L = πλ = π*(100/π) = 100.

Same result as before. Therefore, the maximum area is 100*(200/π) = 20000/π.

Therefore, regardless of the method, I get the same answer.

Therefore, I think that is the correct answer.

But let me check if there's any possible mistake in interpreting the problem.

The problem says "a rectangular field is enclosed in an athletic track". The track consists of two edges of the field and two semicircles. The length of the track is 400 meters.

Wait, maybe "length of the track" refers to the total length around the track, which would be the perimeter of the track. Which we have considered as 2L + πW.

Alternatively, maybe the length of the track refers to the length of the straightaways? But that would be ambiguous. However, given the Asymptote code, which shows the track as having two straight sides and two semicircular sides, and the problem says the length of the track is 400 meters, which is likely the total length around the track.

Therefore, I think my interpretation is correct.

Therefore, the maximum area is 20000/π square meters.

But let me check if the problem is from a standard source. Wait, 400 meters is the standard length for a running track. In reality, a standard 400m track has two straights and two semicircles. The length of the track is 400 meters.

In reality, the maximum area of the field (the inner part of the track) would be when the track is designed such that the area is maximized.

But according to our calculation, the maximum area is 20000/π ≈ 6366.197 m².

But let me check with real-world data. For example, in a standard 400m track, the radius of the semicircular parts is about 36.5 meters. Let me check.

The total length of the track is 2L + 2πr = 400.

But in reality, the length of the straightaways is usually 84.39 meters, and the radius of the semicircular parts is 36.5 meters.

Let me check: 2*84.39 + 2π*36.5 ≈ 168.78 + 2*3.1416*36.5 ≈ 168.78 + 230 ≈ 398.78 meters, which is approximately 400 meters.

So in reality, the area of the field (the inner part) would be length times width. The length is 84.39 meters, and the width is 2*radius = 73 meters. Therefore, area is 84.39*73 ≈ 6160 m².

Which is close to our calculated value of 20000/π ≈ 6366 m².

So our answer is a bit larger than the real-world example. But that's because in reality, the track has lanes, and the inner part is not the maximum area. However, in our problem, we are to find the maximum area of the field enclosed in the track, given that the total length of the track is 400 meters.

Therefore, our answer of 20000/π is correct.

Therefore, I think that is the answer.

**Final Answer**
The largest possible area of the field is \boxed{\dfrac{20000}{\pi}} square meters.

To determine the **largest possible area** of a rectangular field enclosed in an athletic track, we are given that the **total length of the track is 400 meters**. The track consists of:

- Two **straight edges** of the field (i.e., the length of the rectangle),
- Two **semicircular ends** (i.e., the width of the rectangle).

---

### Step 1: Define Variables

Let:
- $ L $ be the **length** of the field (in meters),
- $ W $ be the **width** of the field (in meters).

The **total length of the track** is the sum of:
- The two straight sides: $ 2L $,
- The two semicircular ends: since the width $ W $ is the **diameter** of the semicircles, the **total curved length** is $ \pi W $ (equivalent to the circumference of a full circle with diameter $ W $).

So, the total length of the track is:
$$
2L + \pi W = 400
$$

---

### Step 2: Express Area in Terms of One Variable

We are to **maximize the area** of the field:
$$
A = L \cdot W
$$

From the constraint $ 2L + \pi W = 400 $, solve for $ L $:
$$
L = \frac{400 - \pi W}{2}
$$

Substitute into the area formula:
$$
A = \left( \frac{400 - \pi W}{2} \right) \cdot W = \frac{400W - \pi W^2}{2}
$$

---

### Step 3: Maximize the Area

Let:
$$
A(W) = \frac{400W - \pi W^2}{2}
$$

Take the derivative with respect to $ W $:
$$
\frac{dA}{dW} = \frac{400 - 2\pi W}{2} = 200 - \pi W
$$

Set the derivative equal to zero to find the critical point:
$$
200 - \pi W = 0 \Rightarrow W = \frac{200}{\pi}
$$

Now substitute back to find $ L $:
$$
L = \frac{400 - \pi \cdot \frac{200}{\pi}}{2} = \frac{400 - 200}{2} = 100
$$

---

### Step 4: Compute the Maximum Area

$$
A = L \cdot W = 100 \cdot \frac{200}{\pi} = \frac{20000}{\pi}
$$

---

### Final Answer

$$
\boxed{\dfrac{20000}{\pi}}
$$

This is the **largest possible area** of the rectangular field enclosed in the 400-meter athletic track.

Solution:
Let the width of the rectangle be $w,$ and let the radius of each semicircle be $r.$

[asy]
unitsize(1 cm);

filldraw((0,0)--(3,0)--(3,2)--(0,2)--cycle,lightgreen);
draw((0,0)--(3,0),linewidth(2*bp));
draw((0,2)--(3,2),linewidth(2*bp));
draw(arc((3,1),1,-90,90),linewidth(2*bp));
draw(arc((0,1),1,90,270),linewidth(2*bp));

label("$w$", (1.5,0), S);
label("$r$", (3,1/2), E);
dot((3,1));
[/asy]

Then the length of the track is $2w + 2 \pi r = 400,$ so $w + \pi r = 200.$  By AM-GM,
\[200 = w + \pi r \ge 2 \sqrt{w \pi r},\]so $\sqrt{w \pi r} \le 100.$  Then $w \pi r \le 10000,$ so
\[wr \le \frac{10000}{\pi}.\]Then the area of the field, $2wr,$ must satisfy
\[2wr \le \frac{20000}{\pi}.\]Equality occurs when $w = 100$ and $r = \frac{100}{\pi},$ so the largest possible area is $\boxed{\frac{20000}{\pi}}.$


Explanation:
Okay, let's see. Kathy has $300 in the bank and she wants to withdraw half of it in British pounds and half in euros. Then we need to find out how many more euros than pounds she will have. Hmm, okay. Let me break this down step by step.

First, she wants to withdraw half of her $300 in pounds and half in euros. So, half of $300 is $150. That means she wants to convert $150 into pounds and $150 into euros. Wait, but the question says "withdraw half of it in British pounds and half of it in euros." So does that mean she's withdrawing half the amount in pounds and half in euros? Or is it half the amount in terms of the currency? Wait, maybe I need to clarify.

Wait, the problem says she has 300 USD in the bank. She wants to withdraw half of it in British pounds and half of it in euros. So, does that mean she's withdrawing half of the total amount (which is $150) in pounds and the other half ($150) in euros? But how does that work? Because if she's withdrawing in different currencies, the amount in pounds and euros would depend on the exchange rates.

Wait, maybe she wants to convert half of her USD into pounds and half into euros. So, half of $300 is $150. So she converts $150 to pounds and $150 to euros. Then, we need to find out how much that is in pounds and euros, and then find the difference between euros and pounds.

Yes, that seems to make sense. So let me confirm. The problem says "withdraw half of it in British pounds and half of it in euros." So "it" refers to the 300 USD. So half of it is $150. So she's converting $150 to pounds and $150 to euros. Then, the question is how many more euros than pounds will she have. So, after converting, we need to calculate the amount in euros and pounds, then subtract pounds from euros to find the difference.

Okay, so let's start with converting $150 to pounds. The exchange rate is 1 pound = 1.64 USD. So, to find out how many pounds she gets for $150, we divide the USD amount by the exchange rate. So pounds = USD / 1.64. So that would be 150 / 1.64.

Similarly, for euros, the exchange rate is 1 euro = 1.32 USD. So euros = USD / 1.32. So that would be 150 / 1.32.

Once we have both amounts, we subtract pounds from euros to find the difference. Then round to the nearest whole number.

Let me compute these values step by step.

First, converting $150 to pounds:

150 divided by 1.64. Let me calculate that. Let me get a calculator or do it manually. Let me try to compute 150 / 1.64.

Well, 1.64 times 90 is 147.6, because 1.64 * 90 = 1.64 * 9 * 10 = 14.76 * 10 = 147.6. Then 150 - 147.6 = 2.4. So 2.4 / 1.64 is approximately 1.463. So total pounds would be approximately 90 + 1.463 = 91.463 pounds. Let me check with a calculator approach.

Alternatively, 1.64 * 91 = 1.64 * 90 + 1.64 * 1 = 147.6 + 1.64 = 149.24. Then 150 - 149.24 = 0.76. So 0.76 / 1.64 ≈ 0.463. So total pounds ≈ 91.463. So approximately 91.46 pounds. Let me check with decimal division.

Alternatively, 150 / 1.64. Let me write this as 150 ÷ 1.64. Multiply numerator and denominator by 100 to eliminate decimals: 15000 ÷ 164. Let me compute that.

Divide 15000 by 164. Let me see how many times 164 goes into 15000.

First, 164 * 90 = 14760. Subtract that from 15000: 15000 - 14760 = 240. Then, 164 goes into 240 once (164), remainder 76. Then bring down a zero: 760. 164 * 4 = 656. Subtract from 760: 104. Bring down a zero: 1040. 164 * 6 = 984. Subtract: 1040 - 984 = 56. Bring down a zero: 560. 164 * 3 = 492. Subtract: 560 - 492 = 68. Bring down a zero: 680. 164 * 4 = 656. Subtract: 680 - 656 = 24. So so far, we have 91.463... So approximately 91.46 pounds. So rounding to two decimal places, 91.46 pounds. But since we need to round to the nearest whole number at the end, maybe we can keep more decimals for accuracy.

Alternatively, maybe I can use calculator-like steps. Let me check with approximate decimal values.

Alternatively, maybe I can use fractions. But perhaps it's easier to just proceed with decimal approximations.

So pounds ≈ 91.463.

Now, converting $150 to euros. The exchange rate is 1 euro = 1.32 USD. So euros = 150 / 1.32.

Again, let me compute that. Let me do 150 divided by 1.32.

First, note that 1.32 * 100 = 132. So 150 / 1.32 = (150 * 100) / 132 = 15000 / 132. Let me compute that.

Divide 15000 by 132. Let me see how many times 132 goes into 15000.

First, 132 * 100 = 13200. Subtract that from 15000: 15000 - 13200 = 1800.

Now, 132 * 13 = 1716. Subtract that from 1800: 1800 - 1716 = 84.

Bring down a zero: 840. 132 * 6 = 792. Subtract: 840 - 792 = 48.

Bring down a zero: 480. 132 * 3 = 396. Subtract: 480 - 396 = 84.

Bring down a zero: 840. Wait, this is repeating. So the decimal is 113.6363..., because 100 + 13.6363... So total is 113.6363... euros.

Wait, let me check again. Wait, 132 * 113 = ?

Wait, maybe I should do this step by step. Let me try again.

Original division: 15000 / 132.

First, divide numerator and denominator by 12 to simplify? Let me see. 132 divided by 12 is 11. 15000 divided by 12 is 1250. So now it's 1250 / 11. Which is approximately 113.6363... So yes, that's correct. So 1250 divided by 11 is 113.6363...

So euros ≈ 113.6363...

So approximately 113.64 euros.

So now, we have approximately 91.46 pounds and 113.64 euros.

The question asks how many more euros than pounds she will have. So we need to compute euros - pounds.

So 113.64 - 91.46 = ?

Let me compute that. 113.64 - 91.46. Subtract the whole numbers: 113 - 91 = 22. Subtract the decimals: 0.64 - 0.46 = 0.18. So total is 22.18. So approximately 22.18 euros more.

But we need to round to the nearest whole number. So 22.18 is approximately 22. So the answer would be 22.

Wait, but let me check if my approximated values for pounds and euros are accurate enough. Because if I rounded too early, maybe there's a slight error.

Let me recast the problem with more precise numbers.

First, pounds: 150 / 1.64.

Let me compute 150 / 1.64 exactly. Let me use decimal division.

1.64 × 91 = 1.64 × 90 + 1.64 × 1 = 147.6 + 1.64 = 149.24

Then, 150 - 149.24 = 0.76

So, 0.76 / 1.64 = 0.463...

So total pounds is 91.463...

Similarly, euros: 150 / 1.32 = 113.636...

So euros - pounds = 113.636... - 91.463... = ?

Let me compute that:

113.636... - 91.463... = (113 - 91) + (0.636... - 0.463...) = 22 + 0.173... ≈ 22.173...

So approximately 22.173, which rounds to 22 when rounded to the nearest whole number.

But let me check if there's a more precise way. Maybe I should carry out the exact calculation with more decimal places.

Alternatively, perhaps I can compute the exact difference without rounding intermediate steps.

Let me try that.

Let me compute euros - pounds = (150 / 1.32) - (150 / 1.64)

Factor out 150: 150*(1/1.32 - 1/1.64)

Compute 1/1.32 and 1/1.64.

First, 1/1.32. Let me compute that. 1 divided by 1.32.

Well, 1.32 × 0.75 = 0.99. So 0.75 gives 0.99. The difference is 1 - 0.99 = 0.01. So 0.75 + (0.01 / 1.32). 0.01 / 1.32 ≈ 0.007575... So total is approximately 0.757575...

Similarly, 1/1.64. Let me compute that. 1 divided by 1.64.

Well, 1.64 × 0.6 = 0.984. 1 - 0.984 = 0.016. So 0.6 + (0.016 / 1.64). 0.016 / 1.64 ≈ 0.009756... So total is approximately 0.609756...

Therefore, 1/1.32 - 1/1.64 ≈ 0.757575... - 0.609756... ≈ 0.147819...

Then multiply by 150: 150 × 0.147819 ≈ ?

Compute 150 × 0.147819:

First, 100 × 0.147819 = 14.7819

50 × 0.147819 = 7.39095

Total ≈ 14.7819 + 7.39095 ≈ 22.17285

So approximately 22.17285, which is about 22.17, so rounds to 22. So same as before.

Therefore, the answer is 22.

But let me check once more with exact fractions to see if there's any discrepancy.

Alternatively, let me compute 1/1.32 - 1/1.64.

Convert to fractions:

1.32 is 132/100 = 33/25.

So 1/(33/25) = 25/33 ≈ 0.757575...

Similarly, 1.64 is 164/100 = 41/25.5? Wait, 1.64 is 164/100 = 41/25? Wait 164 divided by 4 is 41, 100 divided by 4 is 25. Yes, so 1.64 = 41/25. Therefore, 1/(41/25) = 25/41 ≈ 0.609756...

Therefore, the difference is 25/33 - 25/41 = 25*(1/33 - 1/41) = 25*( (41 - 33)/(33*41) ) = 25*(8/(1353)) = 200/1353 ≈ ?

Compute 200 divided by 1353. Let me compute that.

Well, 1353 × 0.147 ≈ 1353 × 0.1 + 1353 × 0.04 + 1353 × 0.007

= 135.3 + 54.12 + 9.471 ≈ 135.3 + 54.12 = 189.42 + 9.471 ≈ 198.891

Which is very close to 200. So 0.147 gives approximately 198.89, so 200 / 1353 ≈ 0.1478. Therefore, multiplying by 150:

0.1478 × 150 ≈ 22.17, same as before. So yes, 22.17, rounds to 22.

Therefore, the answer is 22.

But let me check once again with calculator-like precision.

Alternatively, maybe I can use decimal approximations for 1/1.32 and 1/1.64.

Let me compute 1/1.32:

1.32 × 0.75 = 0.99, as before. So 0.75 + (0.01 / 1.32). 0.01 / 1.32 = 0.00757575...

So 1/1.32 ≈ 0.757575...

Similarly, 1/1.64:

Let me do long division for 1 ÷ 1.64.

Alternatively, note that 1.64 × 0.6 = 0.984, as before. Then 1 - 0.984 = 0.016. Bring down a zero: 0.160. 1.64 × 0.009 = 0.01476. Subtract from 0.160: 0.160 - 0.01476 = 0.14524. Bring down a zero: 1.4524. 1.64 × 0.008 = 0.1312. Subtract: 1.4524 - 0.1312 = 1.3212. Bring down a zero: 13.212. 1.64 × 0.008 = 0.1312 again? Wait, this seems messy. Maybe it's better to accept that 1/1.64 ≈ 0.609756...

So, 0.757575... - 0.609756... = 0.147819...

Multiply by 150: 0.147819 * 150 = ?

Well, 0.1 * 150 = 15

0.04 * 150 = 6

0.007819 * 150 ≈ 1.17285

So total ≈ 15 + 6 + 1.17285 ≈ 22.17285, same as before. So 22.17, rounds to 22.

Therefore, the answer is 22.

But let me check once more with exact decimal calculations for pounds and euros.

For pounds: 150 / 1.64.

Let me compute 1.64 × 91.46 = ?

First, 1.64 × 90 = 147.6

1.64 × 1.46 = ?

Compute 1.64 × 1 = 1.64

1.64 × 0.4 = 0.656

1.64 × 0.06 = 0.0984

Adding those: 1.64 + 0.656 = 2.296 + 0.0984 = 2.3944

So total 1.64 × 91.46 = 147.6 + 2.3944 = 149.9944, which is approximately $150. So that checks out. Therefore, 91.46 pounds is correct.

Similarly, for euros: 150 / 1.32 ≈ 113.636...

Let me check 1.32 × 113.636 ≈ ?

First, 1.32 × 113 = ?

1.32 × 100 = 132

1.32 × 13 = 17.16

Total: 132 + 17.16 = 149.16

Now, 1.32 × 0.636 ≈ ?

Well, 1.32 × 0.6 = 0.792

1.32 × 0.036 = 0.04752

Total ≈ 0.792 + 0.04752 = 0.83952

So total euros amount: 149.16 + 0.83952 ≈ 149.99952, which is approximately $150. So that checks out. Therefore, 113.636... euros is correct.

Therefore, euros - pounds = 113.636... - 91.46... ≈ 22.17, which rounds to 22.

Therefore, the answer is 22.

But just to make sure, let me check if there's any chance that the question meant something else. For example, maybe she withdraws half the amount in pounds and half in euros, meaning that the total amount withdrawn is half in pounds and half in euros? Wait, but that would be different. Wait, but the original amount is 300 USD. If she wants to withdraw half of it in pounds and half in euros, that would mean she's converting half of the USD to pounds and half to euros. Which is what I did. So that seems correct.

Alternatively, if she wanted to withdraw half the amount in pounds and half in euros, meaning that the total amount withdrawn is 300 USD, with half in pounds and half in euros? But that would be impossible because pounds and euros are different currencies. So that interpretation doesn't make sense. Therefore, the correct interpretation is converting half of the USD to pounds and half to euros.

Therefore, the answer is 22.

**Final Answer**
\boxed{22}

To determine how many more euros than pounds Kathy will have after withdrawing half of her $300 in British pounds and the other half in euros, we follow a step-by-step process using the given exchange rates:

---

### **Step 1: Determine the amount to convert**

Kathy has **$300** in USD. She wants to withdraw **half** of it in **British pounds (GBP)** and **half** in **Euros (EUR)**.

- Half of $300 = $150
- So, she will convert **$150 to GBP** and **$150 to EUR**

---

### **Step 2: Convert USD to GBP**

Given:
- 1 GBP = 1.64 USD
- So, to find how many GBP she gets for $150:

$$
\text{GBP} = \frac{150}{1.64} \approx 91.463
$$

---

### **Step 3: Convert USD to EUR**

Given:
- 1 EUR = 1.32 USD
- So, to find how many EUR she gets for $150:

$$
\text{EUR} = \frac{150}{1.32} \approx 113.636
$$

---

### **Step 4: Find the difference between euros and pounds**

Now, subtract the number of pounds from the number of euros:

$$
\text{Difference} = 113.636 - 91.463 = 22.173
$$

---

### **Step 5: Round to the nearest whole number**

$$
22.173 \approx \boxed{22}
$$

---

### **Final Answer**

$$
\boxed{22}
$$

[Assistant]
Explanation:
Okay, let's see. Kathy has $300 in the bank and she wants to withdraw half of it in British pounds and half in euros. Then we need to find out how many more euros than pounds she will have. Hmm, okay. Let me break this down step by step.

First, she wants to withdraw half of her $300 in pounds and half in euros. So, half of $300 is $150. That means she wants to convert $150 into pounds and $150 into euros. Wait, but the question says "withdraw half of it in British pounds and half of it in euros." So does that mean she's withdrawing half the amount in pounds and half in euros? Or is it half the amount in terms of the currency? Wait, maybe I need to clarify.

Wait, the problem says she has 300 USD in the bank. She wants to withdraw half of it in British pounds and half of it in euros. So, does that mean she's withdrawing half of the total amount (which is $150) in pounds and the other half ($150) in euros? But how does that work? Because if she's withdrawing in different currencies, the amount in pounds and euros would depend on the exchange rates.

Wait, maybe she wants to convert half of her USD into pounds and half into euros. So, half of $300 is $150. So she converts $150 to pounds and $150 to euros. Then, we need to find out how much that is in pounds and euros, and then find the difference between euros and pounds.

Yes, that seems to make sense. So let me confirm. The problem says "withdraw half of it in British pounds and half of it in euros." So "it" refers to the 300 USD. So half of it is $150. So she's converting $150 to pounds and $150 to euros. Then, the question is how many more euros than pounds will she have. So, after converting, we need to calculate the amount in euros and pounds, then subtract pounds from euros to find the difference.

Okay, so let's start with converting $150 to pounds. The exchange rate is 1 pound = 1.64 USD. So, to find out how many pounds she gets for $150, we divide the USD amount by the exchange rate. So pounds = USD / 1.64. So that would be 150 / 1.64.

Similarly, for euros, the exchange rate is 1 euro = 1.32 USD. So euros = USD / 1.32. So that would be 150 / 1.32.

Once we have both amounts, we subtract pounds from euros to find the difference. Then round to the nearest whole number.

Let me compute these values step by step.

First, converting $150 to pounds:

150 divided by 1.64. Let me calculate that. Let me get a calculator or do it manually. Let me try to compute 150 / 1.64.

Well, 1.64 times 90 is 147.6, because 1.64 * 90 = 1.64 * 9 * 10 = 14.76 * 10 = 147.6. Then 150 - 147.6 = 2.4. So 2.4 / 1.64 is approximately 1.463. So total pounds would be approximately 90 + 1.463 = 91.463 pounds. Let me check with a calculator approach.

Alternatively, 1.64 * 91 = 1.64 * 90 + 1.64 * 1 = 147.6 + 1.64 = 149.24. Then 150 - 149.24 = 0.76. So 0.76 / 1.64 ≈ 0.463. So total pounds ≈ 91.463. So approximately 91.46 pounds. Let me check with decimal division.

Alternatively, 150 / 1.64. Let me write this as 150 ÷ 1.64. Multiply numerator and denominator by 100 to eliminate decimals: 15000 ÷ 164. Let me compute that.

Divide 15000 by 164. Let me see how many times 164 goes into 15000.

First, 164 * 90 = 14760. Subtract that from 15000: 15000 - 14760 = 240. Then, 164 goes into 240 once (164), remainder 76. Then bring down a zero: 760. 164 * 4 = 656. Subtract from 760: 104. Bring down a zero: 1040. 164 * 6 = 984. Subtract: 1040 - 984 = 56. Bring down a zero: 560. 164 * 3 = 492. Subtract: 560 - 492 = 68. Bring down a zero: 680. 164 * 4 = 656. Subtract: 680 - 656 = 24. So so far, we have 91.463... So approximately 91.46 pounds. So rounding to two decimal places, 91.46 pounds. But since we need to round to the nearest whole number at the end, maybe we can keep more decimals for accuracy.

Alternatively, maybe I can use calculator-like steps. Let me check with approximate decimal values.

Alternatively, maybe I can use fractions. But perhaps it's easier to just proceed with decimal approximations.

So pounds ≈ 91.463.

Now, converting $150 to euros. The exchange rate is 1 euro = 1.32 USD. So euros = 150 / 1.32.

Again, let me compute that. Let me do 150 divided by 1.32.

First, note that 1.32 * 100 = 132. So 150 / 1.32 = (150 * 100) / 132 = 15000 / 132. Let me compute that.

Divide 15000 by 132. Let me see how many times 132 goes into 15000.

First, 132 * 100 = 13200. Subtract that from 15000: 15000 - 13200 = 1800.

Now, 132 * 13 = 1716. Subtract that from 1800: 1800 - 1716 = 84.

Bring down a zero: 840. 132 * 6 = 792. Subtract: 840 - 792 = 48.

Bring down a zero: 480. 132 * 3 = 396. Subtract: 480 - 396 = 84.

Bring down a zero: 840. Wait, this is repeating. So the decimal is 113.6363..., because 100 + 13.6363... So total is 113.6363... euros.

Wait, let me check again. Wait, 132 * 113 = ?

Wait, maybe I should do this step by step. Let me try again.

Original division: 15000 / 132.

First, divide numerator and denominator by 12 to simplify? Let me see. 132 divided by 12 is 11. 15000 divided by 12 is 1250. So now it's 1250 / 11. Which is approximately 113.6363... So yes, that's correct. So 1250 divided by 11 is 113.6363...

So euros ≈ 113.6363...

So approximately 113.64 euros.

So now, we have approximately 91.46 pounds and 113.64 euros.

The question asks how many more euros than pounds she will have. So we need to compute euros - pounds.

So 113.64 - 91.46 = ?

Let me compute that. 113.64 - 91.46. Subtract the whole numbers: 113 - 91 = 22. Subtract the decimals: 0.64 - 0.46 = 0.18. So total is 22.18. So approximately 22.18 euros more.

But we need to round to the nearest whole number. So 22.18 is approximately 22. So the answer would be 22.

Wait, but let me check if my approximated values for pounds and euros are accurate enough. Because if I rounded too early, maybe there's a slight error.

Let me recast the problem with more precise numbers.

First, pounds: 150 / 1.64.

Let me compute 150 / 1.64 exactly. Let me use decimal division.

1.64 × 91 = 1.64 × 90 + 1.64 × 1 = 147.6 + 1.64 = 149.24

Then, 150 - 149.24 = 0.76

So, 0.76 / 1.64 = 0.463...

So total pounds is 91.463...

Similarly, euros: 150 / 1.32 = 113.636...

So euros - pounds = 113.636... - 91.463... = ?

Let me compute that:

113.636... - 91.463... = (113 - 91) + (0.636... - 0.463...) = 22 + 0.173... ≈ 22.173...

So approximately 22.173, which rounds to 22 when rounded to the nearest whole number.

But let me check if there's a more precise way. Maybe I should carry out the exact calculation with more decimal places.

Alternatively, perhaps I can compute the exact difference without rounding intermediate steps.

Let me try that.

Let me compute euros - pounds = (150 / 1.32) - (150 / 1.64)

Factor out 150: 150*(1/1.32 - 1/1.64)

Compute 1/1.32 and 1/1.64.

First, 1/1.32. Let me compute that. 1 divided by 1.32.

Well, 1.32 × 0.75 = 0.99. So 0.75 gives 0.99. The difference is 1 - 0.99 = 0.01. So 0.75 + (0.01 / 1.32). 0.01 / 1.32 ≈ 0.007575... So total is approximately 0.757575...

Similarly, 1/1.64. Let me compute that. 1 divided by 1.64.

Well, 1.64 × 0.6 = 0.984. 1 - 0.984 = 0.016. So 0.6 + (0.016 / 1.64). 0.016 / 1.64 ≈ 0.009756... So total is approximately 0.609756...

Therefore, 1/1.32 - 1/1.64 ≈ 0.757575... - 0.609756... ≈ 0.147819...

Then multiply by 150: 150 × 0.147819 ≈ ?

Compute 150 × 0.147819:

First, 100 × 0.147819 = 14.7819

50 × 0.147819 = 7.39095

Total ≈ 14.7819 + 7.39095 ≈ 22.17285

So approximately 22.17285, which is about 22.17, so rounds to 22. So same as before.

Therefore, the answer is 22.

But let me check once more with exact fractions to see if there's any discrepancy.

Alternatively, let me compute 1/1.32 - 1/1.64.

Convert to fractions:

1.32 is 132/100 = 33/25.

So 1/(33/25) = 25/33 ≈ 0.757575...

Similarly, 1.64 is 164/100 = 41/25.5? Wait, 1.64 is 164/100 = 41/25? Wait 164 divided by 4 is 41, 100 divided by 4 is 25. Yes, so 1.64 = 41/25. Therefore, 1/(41/25) = 25/41 ≈ 0.609756...

Therefore, the difference is 25/33 - 25/41 = 25*(1/33 - 1/41) = 25*( (41 - 33)/(33*41) ) = 25*(8/(1353)) = 200/1353 ≈ ?

Compute 200 divided by 1353. Let me compute that.

Well, 1353 × 0.147 ≈ 1353 × 0.1 + 1353 × 0.04 + 1353 × 0.007

= 135.3 + 54.12 + 9.471 ≈ 135.3 + 54.12 = 189.42 + 9.471 ≈ 198.891

Which is very close to 200. So 0.147 gives approximately 198.89, so 200 / 1353 ≈ 0.1478. Therefore, multiplying by 150:

0.1478 × 150 ≈ 22.17, same as before. So yes, 22.17, rounds to 22.

Therefore, the answer is 22.

But let me check once again with calculator-like precision.

Alternatively, maybe I can use decimal approximations for 1/1.32 and 1/1.64.

Let me compute 1/1.32:

1.32 × 0.75 = 0.99, as before. So 0.75 + (0.01 / 1.32). 0.01 / 1.32 = 0.00757575...

So 1/1.32 ≈ 0.757575...

Similarly, 1/1.64:

Let me do long division for 1 ÷ 1.64.

Alternatively, note that 1.64 × 0.6 = 0.984, as before. Then 1 - 0.984 = 0.016. Bring down a zero: 0.160. 1.64 × 0.009 = 0.01476. Subtract from 0.160: 0.160 - 0.01476 = 0.14524. Bring down a zero: 1.4524. 1.64 × 0.008 = 0.1312. Subtract: 1.4524 - 0.1312 = 1.3212. Bring down a zero: 13.212. 1.64 × 0.008 = 0.1312 again? Wait, this seems messy. Maybe it's better to accept that 1/1.64 ≈ 0.609756...

So, 0.757575... - 0.609756... = 0.147819...

Multiply by 150: 0.147819 * 150 = ?

Well, 0.1 * 150 = 15

0.04 * 150 = 6

0.007819 * 150 ≈ 1.17285

So total ≈ 15 + 6 + 1.17285 ≈ 22.17285, same as before. So 22.17, rounds to 22.

Therefore, the answer is 22.

But let me check once more with exact decimal calculations for pounds and euros.

For pounds: 150 / 1.64.

Let me compute 1.64 × 91.46 = ?

First, 1.64 × 90 = 147.6

1.64 × 1.46 = ?

Compute 1.64 × 1 = 1.64

1.64 × 0.4 = 0.656

1.64 × 0.06 = 0.0984

Adding those: 1.64 + 0.656 = 2.296 + 0.0984 = 2.3944

So total 1.64 × 91.46 = 147.6 + 2.3944 = 149.9944, which is approximately $150. So that checks out. Therefore, 91.46 pounds is correct.

Similarly, for euros: 150 / 1.32 ≈ 113.636...

Let me check 1.32 × 113.636 ≈ ?

First, 1.32 × 113 = ?

1.32 × 100 = 132

1.32 × 13 = 17.16

Total: 132 + 17.16 = 149.16

Now, 1.32 × 0.636 ≈ ?

Well, 1.32 × 0.6 = 0.792

1.32 × 0.036 = 0.04752

Total ≈ 0.792 + 0.04752 = 0.83952

So total euros amount: 149.16 + 0.83952 ≈ 149.99952, which is approximately $150. So that checks out. Therefore, 113.636... euros is correct.

Therefore, euros - pounds = 113.636... - 91.46... ≈ 22.17, which rounds to 22.

Therefore, the answer is 22.

But just to make sure, let me check if there's any chance that the question meant something else. For example, maybe she withdraws half the amount in pounds and half in euros, meaning that the total amount withdrawn is half in pounds and half in euros? Wait, but that would be different. Wait, but the original amount is 300 USD. If she wants to withdraw half of it in pounds and half in euros, that would mean she's converting half of the USD to pounds and half to euros. Which is what I did. So that seems correct.

Alternatively, if she wanted to withdraw half the amount in pounds and half in euros, meaning that the total amount withdrawn is 300 USD, with half in pounds and half in euros? But that would be impossible because pounds and euros are different currencies. So that interpretation doesn't make sense. Therefore, the correct interpretation is converting half of the USD to pounds and half to euros.

Therefore, the answer is 22.

**Final Answer**
\boxed{22}

To determine how many more euros than pounds Kathy will have after withdrawing half of her $300 in British pounds and the other half in euros, we follow a step-by-step process using the given exchange rates:

---

### **Step 1: Determine the amount to convert**

Kathy has **$300** in USD. She wants to withdraw **half** of it in **British pounds (GBP)** and **half** in **Euros (EUR)**.

- Half of $300 = $150
- So, she will convert **$150 to GBP** and **$150 to EUR**

---

### **Step 2: Convert USD to GBP**

Given:
- 1 GBP = 1.64 USD
- So, to find how many GBP she gets for $150:

$$
\text{GBP} = \frac{150}{1.64} \approx 91.463
$$

---

### **Step 3: Convert USD to EUR**

Given:
- 1 EUR = 1.32 USD
- So, to find how many EUR she gets for $150:

$$
\text{EUR} = \frac{150}{1.32} \approx 113.636
$$

---

### **Step 4: Find the difference between euros and pounds**

Now, subtract the number of pounds from the number of euros:

$$
\text{Difference} = 113.636 - 91.463 = 22.173
$$

---

### **Step 5: Round to the nearest whole number**

$$
22.173 \approx \boxed{22}
$$

---

### **Final Answer**

$$
\boxed{22}
$$

Solution: After Kathy has exchanged half of her money ($300\text{ USD}\div 2 = 150\text{ USD}$) into pounds, she will have $150\text{ USD}\times\frac{1\text{ pound}}{1.64 \text{ USD}}\approx 91.46 \text{ pounds}$. After exchanging the other half of her money into euros, she will have $150\text{ USD} \times\frac{1\text{ euro}}{1.32 \text{ USD}}\approx 113.64\text{ euros}$. Subtracting these two values, we have $113.64-91.46=22.18$. Since the question calls for the nearest whole number, we round 22.18 to the final answer of $\boxed{22}$.


Explanation:
Okay, so I need to find the smallest real number x in the domain of the function g(x) = sqrt[(x - 3)^2 - (x - 8)^2]. Hmm, let me think about how to approach this.

First, I remember that for a square root function to be defined (since we're dealing with real numbers), the expression inside the square root must be non-negative. That means whatever is under the square root, which is (x - 3)^2 - (x - 8)^2, has to be greater than or equal to zero. So my first step should be to figure out for which x values this expression is non-negative. Once I have that domain, I can then find the smallest real number in that domain.

Alright, let me start by simplifying the expression inside the square root. Let me compute (x - 3)^2 - (x - 8)^2. Maybe expanding both squares will help.

Let me recall that (a - b)^2 = a^2 - 2ab + b^2. So applying that to both terms:

First term: (x - 3)^2 = x^2 - 6x + 9

Second term: (x - 8)^2 = x^2 - 16x + 64

So subtracting the second term from the first term:

(x^2 - 6x + 9) - (x^2 - 16x + 64)

Let me distribute the negative sign to the second term:

x^2 - 6x + 9 - x^2 + 16x - 64

Now combine like terms:

x^2 - x^2 cancels out, so we have (-6x + 16x) + (9 - 64)

Which is 10x - 55.

So the expression inside the square root simplifies to 10x - 55. Therefore, the original function g(x) can be rewritten as sqrt(10x - 55).

Wait, so that simplifies things a lot. So the domain of g(x) is all real numbers x such that 10x - 55 ≥ 0. Let me write that down:

10x - 55 ≥ 0

Solving for x:

10x ≥ 55

Divide both sides by 10:

x ≥ 55/10

Which simplifies to x ≥ 11/2 or x ≥ 5.5.

Therefore, the domain of g(x) is [5.5, ∞). Therefore, the smallest real number in the domain is 5.5, which is 11/2.

Wait, but let me check my steps again to make sure I didn't make a mistake. Because sometimes when simplifying expressions, especially with squares, there might be errors. Let me verify the algebra again.

Original expression: (x - 3)^2 - (x - 8)^2.

Expanding both squares:

First square: (x - 3)^2 = x² - 6x + 9

Second square: (x - 8)^2 = x² - 16x + 64

Subtracting the second square from the first:

(x² - 6x + 9) - (x² - 16x + 64) = x² - 6x + 9 - x² + 16x - 64.

Yes, the x² terms cancel. Then -6x + 16x is 10x. Then 9 - 64 is -55. So 10x - 55. That seems correct.

So the expression under the square root is 10x - 55. Therefore, 10x - 55 ≥ 0 => x ≥ 5.5.

Therefore, the domain is [5.5, ∞). Therefore, the smallest real number in the domain is 5.5, which is 11/2.

But let me think again. Is there a possibility that the original expression could have been simplified differently? For example, sometimes there are alternative ways to factor expressions. Let me try another approach.

Another way to compute (x - 3)^2 - (x - 8)^2 is to use the difference of squares formula: a² - b² = (a - b)(a + b).

Let me set a = (x - 3) and b = (x - 8). Then:

a² - b² = (a - b)(a + b) = [(x - 3) - (x - 8)] * [(x - 3) + (x - 8)]

Compute each part:

First, (x - 3) - (x - 8) = x - 3 - x + 8 = (x - x) + (-3 + 8) = 0 + 5 = 5

Second, (x - 3) + (x - 8) = x - 3 + x - 8 = 2x - 11

Therefore, the entire expression becomes 5*(2x - 11) = 10x - 55.

Same result as before. So that confirms that the expression inside the square root simplifies to 10x - 55. Therefore, the domain is indeed x ≥ 5.5.

Therefore, the smallest real number in the domain is 5.5, which is 11/2.

But let me check if there's any other constraints. For example, sometimes when you have square roots, there might be more to it, but in this case, since we simplified it to sqrt(10x - 55), the only constraint is that 10x - 55 ≥ 0.

So, unless there's some other hidden constraint, but I don't think so. The original function is defined as long as the expression inside the square root is non-negative, which we have already considered.

Therefore, the domain is [11/2, ∞), so the smallest real number in the domain is 11/2.

But just to be thorough, let me plug in x = 11/2 into the original expression to check if it's valid.

Original expression inside sqrt: (x - 3)^2 - (x - 8)^2. Let me compute that when x = 11/2.

First, compute x - 3: 11/2 - 3 = 11/2 - 6/2 = 5/2

Then, (x - 3)^2 = (5/2)^2 = 25/4

Next, x - 8: 11/2 - 8 = 11/2 - 16/2 = -5/2

Then, (x - 8)^2 = (-5/2)^2 = 25/4

Therefore, the expression inside the sqrt is 25/4 - 25/4 = 0.

Which is okay, because sqrt(0) is 0, which is a real number. Therefore, x = 11/2 is indeed in the domain.

If I try x slightly less than 11/2, say x = 5.4, then 10x - 55 = 54 - 55 = -1, which is negative, so sqrt(-1) is not real. Therefore, x must be at least 11/2.

Therefore, the conclusion is that the smallest real number in the domain is 11/2, which is 5.5.

But let me check if there's another way someone might approach this problem and perhaps make a mistake. For example, someone might think that since there are two squares, maybe there's a different domain? But no, because we simplified it correctly.

Alternatively, someone might think that the expression inside the square root is a quadratic, and maybe they would try to solve the inequality (x - 3)^2 - (x - 8)^2 ≥ 0, but as we saw, that simplifies to 10x - 55 ≥ 0, which is linear, not quadratic.

So, there's no quadratic here. Therefore, the domain is straightforward.

Another check: Let me consider the original function. If I graph the function g(x) = sqrt(10x - 55), it's a square root function shifted to the right by 5.5 units. The domain is x ≥ 5.5, so the smallest x is 5.5.

Alternatively, if I didn't simplify the expression, and tried to think about the original expression (x - 3)^2 - (x - 8)^2, maybe someone might think that it's a quadratic and try to find where it's non-negative. But as we saw, it's linear.

So, in conclusion, the domain is [11/2, ∞), so the smallest real number in the domain is 11/2.

Therefore, the answer should be 11/2, which is 5.5.

But let me check once more for errors. Let me take another value in the domain, say x = 6. Then compute the original expression:

(x - 3)^2 - (x - 8)^2 = (6 - 3)^2 - (6 - 8)^2 = 3^2 - (-2)^2 = 9 - 4 = 5. Which is positive, so sqrt(5) is real.

If x = 5.5, then as before, it's 0. If x = 5, then 10x - 55 = 50 - 55 = -5, which is negative. So indeed, 5.5 is the boundary.

Therefore, I think my answer is correct.

**Final Answer**
The smallest real number $ x $ in the domain of $ g(x) $ is \boxed{\dfrac{11}{2}}.

To determine the smallest real number $ x $ in the domain of the function
$$
g(x) = \sqrt{(x - 3)^2 - (x - 8)^2},
$$
we must ensure that the expression inside the square root is **non-negative**, since the square root of a negative number is not defined in the set of real numbers.

---

### Step 1: Simplify the Expression Inside the Square Root

We begin by simplifying the expression:
$$
(x - 3)^2 - (x - 8)^2.
$$

Using the identity $ a^2 - b^2 = (a - b)(a + b) $, we let:
- $ a = x - 3 $
- $ b = x - 8 $

Then:
$$
(x - 3)^2 - (x - 8)^2 = [(x - 3) - (x - 8)][(x - 3) + (x - 8)].
$$

Simplify each factor:
- $ (x - 3) - (x - 8) = x - 3 - x + 8 = 5 $
- $ (x - 3) + (x - 8) = x - 3 + x - 8 = 2x - 11 $

So the expression becomes:
$$
5(2x - 11) = 10x - 55.
$$

---

### Step 2: Determine the Domain

Now, the function becomes:
$$
g(x) = \sqrt{10x - 55}.
$$

For the square root to be defined in the real numbers, the expression inside must be **non-negative**:
$$
10x - 55 \geq 0.
$$

Solving this inequality:
$$
10x \geq 55 \quad \Rightarrow \quad x \geq \frac{55}{10} = \frac{11}{2}.
$$

Thus, the domain of $ g(x) $ is:
$$
\left[ \frac{11}{2}, \infty \right).
$$

---

### Step 3: Find the Smallest Real Number in the Domain

The smallest real number in the interval $ \left[ \frac{11}{2}, \infty \right) $ is:
$$
\boxed{\dfrac{11}{2}}.
$$

---

### Final Answer

$$
\boxed{\dfrac{11}{2}}
$$

[Assistant]
Explanation:
Okay, so I need to find the smallest real number x in the domain of the function g(x) = sqrt[(x - 3)^2 - (x - 8)^2]. Hmm, let me think about how to approach this.

First, I remember that for a square root function to be defined (since we're dealing with real numbers), the expression inside the square root must be non-negative. That means whatever is under the square root, which is (x - 3)^2 - (x - 8)^2, has to be greater than or equal to zero. So my first step should be to figure out for which x values this expression is non-negative. Once I have that domain, I can then find the smallest real number in that domain.

Alright, let me start by simplifying the expression inside the square root. Let me compute (x - 3)^2 - (x - 8)^2. Maybe expanding both squares will help.

Let me recall that (a - b)^2 = a^2 - 2ab + b^2. So applying that to both terms:

First term: (x - 3)^2 = x^2 - 6x + 9

Second term: (x - 8)^2 = x^2 - 16x + 64

So subtracting the second term from the first term:

(x^2 - 6x + 9) - (x^2 - 16x + 64)

Let me distribute the negative sign to the second term:

x^2 - 6x + 9 - x^2 + 16x - 64

Now combine like terms:

x^2 - x^2 cancels out, so we have (-6x + 16x) + (9 - 64)

Which is 10x - 55.

So the expression inside the square root simplifies to 10x - 55. Therefore, the original function g(x) can be rewritten as sqrt(10x - 55).

Wait, so that simplifies things a lot. So the domain of g(x) is all real numbers x such that 10x - 55 ≥ 0. Let me write that down:

10x - 55 ≥ 0

Solving for x:

10x ≥ 55

Divide both sides by 10:

x ≥ 55/10

Which simplifies to x ≥ 11/2 or x ≥ 5.5.

Therefore, the domain of g(x) is [5.5, ∞). Therefore, the smallest real number in the domain is 5.5, which is 11/2.

Wait, but let me check my steps again to make sure I didn't make a mistake. Because sometimes when simplifying expressions, especially with squares, there might be errors. Let me verify the algebra again.

Original expression: (x - 3)^2 - (x - 8)^2.

Expanding both squares:

First square: (x - 3)^2 = x² - 6x + 9

Second square: (x - 8)^2 = x² - 16x + 64

Subtracting the second square from the first:

(x² - 6x + 9) - (x² - 16x + 64) = x² - 6x + 9 - x² + 16x - 64.

Yes, the x² terms cancel. Then -6x + 16x is 10x. Then 9 - 64 is -55. So 10x - 55. That seems correct.

So the expression under the square root is 10x - 55. Therefore, 10x - 55 ≥ 0 => x ≥ 5.5.

Therefore, the domain is [5.5, ∞). Therefore, the smallest real number in the domain is 5.5, which is 11/2.

But let me think again. Is there a possibility that the original expression could have been simplified differently? For example, sometimes there are alternative ways to factor expressions. Let me try another approach.

Another way to compute (x - 3)^2 - (x - 8)^2 is to use the difference of squares formula: a² - b² = (a - b)(a + b).

Let me set a = (x - 3) and b = (x - 8). Then:

a² - b² = (a - b)(a + b) = [(x - 3) - (x - 8)] * [(x - 3) + (x - 8)]

Compute each part:

First, (x - 3) - (x - 8) = x - 3 - x + 8 = (x - x) + (-3 + 8) = 0 + 5 = 5

Second, (x - 3) + (x - 8) = x - 3 + x - 8 = 2x - 11

Therefore, the entire expression becomes 5*(2x - 11) = 10x - 55.

Same result as before. So that confirms that the expression inside the square root simplifies to 10x - 55. Therefore, the domain is indeed x ≥ 5.5.

Therefore, the smallest real number in the domain is 5.5, which is 11/2.

But let me check if there's any other constraints. For example, sometimes when you have square roots, there might be more to it, but in this case, since we simplified it to sqrt(10x - 55), the only constraint is that 10x - 55 ≥ 0.

So, unless there's some other hidden constraint, but I don't think so. The original function is defined as long as the expression inside the square root is non-negative, which we have already considered.

Therefore, the domain is [11/2, ∞), so the smallest real number in the domain is 11/2.

But just to be thorough, let me plug in x = 11/2 into the original expression to check if it's valid.

Original expression inside sqrt: (x - 3)^2 - (x - 8)^2. Let me compute that when x = 11/2.

First, compute x - 3: 11/2 - 3 = 11/2 - 6/2 = 5/2

Then, (x - 3)^2 = (5/2)^2 = 25/4

Next, x - 8: 11/2 - 8 = 11/2 - 16/2 = -5/2

Then, (x - 8)^2 = (-5/2)^2 = 25/4

Therefore, the expression inside the sqrt is 25/4 - 25/4 = 0.

Which is okay, because sqrt(0) is 0, which is a real number. Therefore, x = 11/2 is indeed in the domain.

If I try x slightly less than 11/2, say x = 5.4, then 10x - 55 = 54 - 55 = -1, which is negative, so sqrt(-1) is not real. Therefore, x must be at least 11/2.

Therefore, the conclusion is that the smallest real number in the domain is 11/2, which is 5.5.

But let me check if there's another way someone might approach this problem and perhaps make a mistake. For example, someone might think that since there are two squares, maybe there's a different domain? But no, because we simplified it correctly.

Alternatively, someone might think that the expression inside the square root is a quadratic, and maybe they would try to solve the inequality (x - 3)^2 - (x - 8)^2 ≥ 0, but as we saw, that simplifies to 10x - 55 ≥ 0, which is linear, not quadratic.

So, there's no quadratic here. Therefore, the domain is straightforward.

Another check: Let me consider the original function. If I graph the function g(x) = sqrt(10x - 55), it's a square root function shifted to the right by 5.5 units. The domain is x ≥ 5.5, so the smallest x is 5.5.

Alternatively, if I didn't simplify the expression, and tried to think about the original expression (x - 3)^2 - (x - 8)^2, maybe someone might think that it's a quadratic and try to find where it's non-negative. But as we saw, it's linear.

So, in conclusion, the domain is [11/2, ∞), so the smallest real number in the domain is 11/2.

Therefore, the answer should be 11/2, which is 5.5.

But let me check once more for errors. Let me take another value in the domain, say x = 6. Then compute the original expression:

(x - 3)^2 - (x - 8)^2 = (6 - 3)^2 - (6 - 8)^2 = 3^2 - (-2)^2 = 9 - 4 = 5. Which is positive, so sqrt(5) is real.

If x = 5.5, then as before, it's 0. If x = 5, then 10x - 55 = 50 - 55 = -5, which is negative. So indeed, 5.5 is the boundary.

Therefore, I think my answer is correct.

**Final Answer**
The smallest real number $ x $ in the domain of $ g(x) $ is \boxed{\dfrac{11}{2}}.

To determine the smallest real number $ x $ in the domain of the function
$$
g(x) = \sqrt{(x - 3)^2 - (x - 8)^2},
$$
we must ensure that the expression inside the square root is **non-negative**, since the square root of a negative number is not defined in the set of real numbers.

---

### Step 1: Simplify the Expression Inside the Square Root

We begin by simplifying the expression:
$$
(x - 3)^2 - (x - 8)^2.
$$

Using the identity $ a^2 - b^2 = (a - b)(a + b) $, we let:
- $ a = x - 3 $
- $ b = x - 8 $

Then:
$$
(x - 3)^2 - (x - 8)^2 = [(x - 3) - (x - 8)][(x - 3) + (x - 8)].
$$

Simplify each factor:
- $ (x - 3) - (x - 8) = x - 3 - x + 8 = 5 $
- $ (x - 3) + (x - 8) = x - 3 + x - 8 = 2x - 11 $

So the expression becomes:
$$
5(2x - 11) = 10x - 55.
$$

---

### Step 2: Determine the Domain

Now, the function becomes:
$$
g(x) = \sqrt{10x - 55}.
$$

For the square root to be defined in the real numbers, the expression inside must be **non-negative**:
$$
10x - 55 \geq 0.
$$

Solving this inequality:
$$
10x \geq 55 \quad \Rightarrow \quad x \geq \frac{55}{10} = \frac{11}{2}.
$$

Thus, the domain of $ g(x) $ is:
$$
\left[ \frac{11}{2}, \infty \right).
$$

---

### Step 3: Find the Smallest Real Number in the Domain

The smallest real number in the interval $ \left[ \frac{11}{2}, \infty \right) $ is:
$$
\boxed{\dfrac{11}{2}}.
$$

---

### Final Answer

$$
\boxed{\dfrac{11}{2}}
$$

Solution:
A real number $x$ is in the domain of $g$ if and only if $$(x-3)^2 - (x-8)^2 \ge 0.$$ Expanding this out and simplifying, we get $$10x - 55\ge 0;$$ the smallest solution is $x=\frac{55}{10}=\boxed{\frac{11}{2}}$.

Alternatively, once we have the quadratic equation $$(x-3)^2 - (x-8)^2 \ge 0,$$ instead of expanding it out, we can observe that $(x-3)^2$ is the square of the distance from $x$ to $3$ on the number line, while $(x-8)^2$ is the square of the distance from $x$ to $8$. Thus, $(x-3)^2-(x-8)^2\ge 0$ is true if $x$ is closer to $8$ than to $3$, which is true if and only if $x\ge \frac{8+3}{2} = \boxed{\frac{11}{2}}$.