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. Let me try to visualize this. The Asymptote code shows a rectangle with two semicircles on the ends, right? So, the track is like a rectangle with semicircular ends, making it similar to a running track. The length of the track is given as 400 meters. Wait, the problem says the track consists of two edges of the field and two semicircles. Hmm, maybe I should clarify that.

Let me think. If the field is rectangular, then the track would be surrounding the field. But the problem says the track consists of two edges of the field and two semicircles. So maybe the track is only the two straight sides of the field and the two semicircular ends? That would make sense. So if you imagine the field as a rectangle, the track is formed by the two longer sides of the rectangle and two semicircles on the ends. So the total length of the track is the sum of the two straight sides and the circumference of the two semicircles.

Wait, but the two semicircles would make a full circle. So if the track's total length is 400 meters, that would be the sum of the lengths of the two straight sides (which are the length of the rectangle) and the circumference of a circle with diameter equal to the width of the rectangle.

Let me write that down. Let me denote the length of the rectangular field as L and the width as W. Then, the two straight sides of the track would be 2L, right? Wait, no. Wait, if the track is surrounding the field, but the problem says the track consists of two edges of the field and two semicircles. So maybe the two edges of the field are the two lengths? So if you have the field as a rectangle, then the track is the two lengths (so 2L) and the two semicircles on the ends. But the two semicircles would be on the width of the field. So each semicircle has a diameter equal to the width W. So the circumference of a full circle with diameter W is πW, so two semicircles would be πW. Therefore, the total length of the track is 2L + πW = 400 meters.

So the problem says the length of the track is 400 meters. So that equation is 2L + πW = 400.

And we need to find the maximum area of the field. The area of the field is length times width, so A = L * W.

So we need to maximize A = L * W, given that 2L + πW = 400.

This is an optimization problem with a constraint. So we can use methods like substitution or Lagrange multipliers. But since it's a single constraint with two variables, substitution seems straightforward.

Let me solve the constraint equation for one variable in terms of the other. Let's solve for L.

2L = 400 - πW

So L = (400 - πW)/2

Then substitute into the area formula:

A = L * W = [(400 - πW)/2] * W = (400W - πW²)/2

So A = 200W - (π/2)W²

Now, to find the maximum area, we can take the derivative of A with respect to W and set it equal to zero.

But wait, since this is a quadratic function in terms of W, it's a downward-opening parabola, so the maximum occurs at the vertex.

Alternatively, I can use calculus. Let me try calculus here.

dA/dW = 200 - πW

Set derivative equal to zero:

200 - πW = 0 => πW = 200 => W = 200 / π

Then, substitute back into the equation for L:

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

So the maximum area occurs when W = 200/π and L = 100.

Then the area A is L * W = 100 * (200 / π) = 20000 / π

Wait, but let me check if this is correct. Alternatively, using the vertex formula for a quadratic.

The quadratic equation is A = - (π/2)W² + 200W.

In general, for a quadratic ax² + bx + c, the vertex is at x = -b/(2a). Here, a = -π/2, b = 200.

So W = -200 / (2*(-π/2)) = -200 / (-π) = 200/π. Yep, same result.

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

But let me check if the problem is referring to the area of the field, which is the rectangle. Wait, but the Asymptote code shows a rectangle with semicircular ends, but maybe the "field" refers to the rectangle? Or is the entire track area? Wait, the problem says "a rectangular field is enclosed in an athletic track". So the field is the rectangle, and the track is around it? Or is the track itself the field?

Wait, the problem says "the track consists of two edges of the field, and two semicircles." So the track is made up of two edges of the field (so the straight sides) and two semicircles. So maybe the field is the rectangle, and the track is surrounding it? Or is the track the field?

Wait, maybe I need to confirm. Let me re-read the problem.

"A rectangular field is enclosed in an athletic track, as shown below. The track consists of two edges of the field, and two semicircles. The length of the track is 400 meters. What is the largest possible area of the field, in square meters?"

So the rectangular field is enclosed by the track. The track consists of two edges of the field and two semicircles. So the track is made of the two straight sides of the field (the length of the rectangle) and two semicircular ends. Therefore, the total length of the track is the two lengths plus the circumference of the circle (since two semicircles make a full circle).

Therefore, as I had earlier, the total perimeter of the track is 2L + πW = 400.

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

So my previous calculation seems right.

But just to be thorough, let me check units. 2L + πW = 400 meters. So if W is in meters, then yes.

So maximum area is 20000 / π. Let me compute that numerically.

But maybe the question wants an exact value, so 20000/π. But maybe I can write it as (200^2)/π. Wait, 200*100 is 20000.

Alternatively, if I think about the formula, maybe there's a different approach.

Wait, another way to think about this is that the track is similar to a standard running track. In reality, the perimeter of a running track is calculated as 2*(length) + 2*(π*(diameter)/2), which is the same as 2L + πW. So that's consistent with our equation.

But perhaps there's a different way to think about maximizing area with a fixed perimeter? Wait, but in this case, the perimeter isn't exactly the same as a regular perimeter. The "track" has a different shape, but the problem defines its length as 400 meters.

Alternatively, perhaps if we think of the entire track as a closed loop. If the track is made up of two straight sides of length L and two semicircular ends with diameter W, then the total length of the track is 2L + πW. So that's given as 400.

Therefore, the problem reduces to maximizing A = L*W with 2L + πW = 400. Which is exactly what I did earlier.

Therefore, solving gives W = 200/π, L = 100, A = 20000/π.

But let me check if this is the maximum. For example, if we take W approaching zero, then L approaches 200, and area approaches zero. If W approaches 400/π, then L approaches zero, and area approaches zero. So the maximum is indeed somewhere in between.

Alternatively, if I think of this as an optimization problem, maybe using Lagrange multipliers. Let me try that.

Let me set up the function to maximize: A = L*W.

Subject to the constraint: 2L + πW = 400.

Using Lagrange multipliers, set up ∇A = λ∇g, where g(L, W) = 2L + πW - 400 = 0.

Compute the gradients:

∇A = (W, L)

∇g = (2, π)

So setting up equations:

W = 2λ

L = πλ

So from the first equation, λ = W/2. Substitute into second equation: L = π*(W/2) => L = (π/2)W.

But from the constraint equation: 2L + πW = 400. Substitute L = (π/2)W:

2*(π/2)W + πW = πW + πW = 2πW = 400 => W = 400/(2π) = 200/π. Then L = (π/2)*(200/π) = 100.

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

So, 20000 divided by pi.

But maybe I need to check if the problem is referring to the area of the track or the field? Wait, the problem says "the largest possible area of the field". The field is the rectangle, so that's L*W. So yes, 20000/pi.

But maybe I need to verify with some numbers. Let me take a concrete example.

Suppose that W = 200/pi ≈ 200/3.1416 ≈ 63.66 meters. Then L = 100 meters.

Then perimeter is 2*100 + π*63.66 ≈ 200 + 200 = 400 meters. Which checks out.

So the area would be 100*63.66 ≈ 6366 square meters.

Alternatively, if I take different values. Suppose W = 50 meters, then 2L + π*50 = 400 => 2L = 400 - 157.08 = 242.92 => L = 121.46. Then area is 121.46*50 ≈ 6073, which is less than 6366.

Another example: let me take W = 70 meters. Then 2L + π*70 ≈ 2L + 219.91 = 400 => 2L ≈ 180.09 => L ≈ 90.045. Area ≈ 90.045 * 70 ≈ 6303. Which is still less than 6366.

So seems like 20000/pi is the maximum.

Alternatively, if I take W = 200/pi ≈ 63.66 meters. Then the area is approx 20000/3.1416 ≈ 6366.2 square meters.

So I think that's correct.

But wait, the problem mentions that the track consists of two edges of the field and two semicircles. Is there a possibility that the semicircles are on the length instead of the width? Wait, no. If the track consists of two edges of the field and two semicircles, the semicircles have to be on the ends of the field. So if the field is a rectangle, the two semicircles would be on the shorter sides, so the diameter of the semicircles would be the width of the field.

Alternatively, if the semicircles were on the length, then the diameter would be length, but that would make the track longer. But given that the track length is 400 meters, which is a standard track length. For example, a standard 400m track has two straightaways and two semicircular ends. In reality, the straightaways are about 84.39 meters each, and the diameter of the semicircular ends is about 73 meters. But of course, the exact numbers depend on the design. However, in our problem, we are to maximize the area of the field (the rectangle).

So, in the calculations, we found that when the width is 200/pi and length is 100, the area is maximized.

Alternatively, if the problem was asking for the area of the entire track, which would be the area of the rectangle plus the area of the circle (since two semicircles make a full circle). But the problem says "the largest possible area of the field", so the field is the rectangle, so the answer should be 20000/pi.

But let me just check if I didn't misinterpret the track's perimeter.

The problem says "the track consists of two edges of the field, and two semicircles". So, if the field is the rectangle, then the track is built along two sides of the field (the length?) and the semicircular ends.

Wait, if the track is surrounding the field, then maybe the two edges of the field are the straight sides of the track. So if the field is a rectangle, then the track would have two straight sides equal to the length of the field and two semicircular ends with diameter equal to the width of the field. Therefore, the total length of the track is 2L + πW.

Alternatively, if the track was built on the field, meaning that the field is inside the track. But the problem says the field is enclosed in the track. So the field is inside. Therefore, the track is the outer perimeter.

Therefore, in that case, the length of the track would be the outer perimeter, which would be 2L + 2*(πW/2) = 2L + πW. Which is again the same equation.

Therefore, the calculation is correct.

Thus, the maximum area of the field is 20000 divided by pi.

But just to confirm, let's think about the standard maximum area for a given perimeter. For a given perimeter, the maximum area is achieved by a circle. However, in this case, the shape is constrained to be a rectangle with semicircular ends, so the maximum area under this constraint is different.

But in this case, we derived that the maximum area is 20000/pi.

Alternatively, if we think of the entire track as a shape with perimeter 400m, but the track has a different shape. However, the problem specifically says that the track consists of two edges of the field and two semicircles, so we have to stick with that.

Therefore, I think my answer is correct.

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

To solve the problem, we interpret the description as follows: a rectangular field is enclosed by a track. The track comprises **two straight sides** of the field (the sides of the rectangle) and **two semicircles** at the ends of the field. Since two semicircles make a full circle, the total length of the track is the sum of the **perimeter of the two straight sides** and the **circumference of a full circle**.

---

### Step 1: Define Variables
Let:
- $ L $ be the **length** of the rectangular field
- $ W $ be the **width** of the rectangular field

The straight segments of the track are the two lengths of the rectangle, so their total is:
$$
2L
$$

The curved part is a full circle with **diameter $ W $**, so its circumference is:
$$
\pi W
$$

Therefore, the total length of the track is:
$$
2L + \pi W = 400 \quad \text{(1)}
$$

The **area of the field**, which is the rectangle, is:
$$
A = L \cdot W \quad \text{(2)}
$$

Our goal is to **maximize** this area, $ A $, under the constraint given by equation (1).

---

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

From equation (1), solve for $ L $:
$$
2L = 400 - \pi W \Rightarrow L = \frac{400 - \pi W}{2}
$$

Substitute into equation (2):
$$
A = L \cdot W = \frac{400 - \pi W}{2} \cdot W = (200 - \frac{\pi}{2} W) \cdot W = 200W - \frac{\pi}{2} W^2
$$

---

### Step 3: Maximize the Area Function

We now analyze the quadratic function:
$$
A(W) = 200W - \frac{\pi}{2} W^2
$$

This is a **downward-opening parabola**, so it has a **maximum** at the vertex. The vertex occurs when:
$$
W = \frac{-b}{2a}
$$

In this case:
- $ a = -\frac{\pi}{2} $
- $ b = 200 $

So:
$$
W = \frac{-200}{2 \cdot (-\frac{\pi}{2})} = \frac{-200}{-\pi} = \frac{200}{\pi}
$$

---

### Step 4: Find the Maximum Area

Now substitute $ W = \frac{200}{\pi} $ back into the expression for $ A $:

First, find $ L $:
$$
L = \frac{400 - \pi W}{2} = \frac{400 - \pi \cdot \frac{200}{\pi}}{2} = \frac{400 - 200}{2} = 100
$$

Then compute the 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, in square meters.

[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. Let me try to visualize this. The Asymptote code shows a rectangle with two semicircles on the ends, right? So, the track is like a rectangle with semicircular ends, making it similar to a running track. The length of the track is given as 400 meters. Wait, the problem says the track consists of two edges of the field and two semicircles. Hmm, maybe I should clarify that.

Let me think. If the field is rectangular, then the track would be surrounding the field. But the problem says the track consists of two edges of the field and two semicircles. So maybe the track is only the two straight sides of the field and the two semicircular ends? That would make sense. So if you imagine the field as a rectangle, the track is formed by the two longer sides of the rectangle and two semicircles on the ends. So the total length of the track is the sum of the two straight sides and the circumference of the two semicircles.

Wait, but the two semicircles would make a full circle. So if the track's total length is 400 meters, that would be the sum of the lengths of the two straight sides (which are the length of the rectangle) and the circumference of a circle with diameter equal to the width of the rectangle.

Let me write that down. Let me denote the length of the rectangular field as L and the width as W. Then, the two straight sides of the track would be 2L, right? Wait, no. Wait, if the track is surrounding the field, but the problem says the track consists of two edges of the field and two semicircles. So maybe the two edges of the field are the two lengths? So if you have the field as a rectangle, then the track is the two lengths (so 2L) and the two semicircles on the ends. But the two semicircles would be on the width of the field. So each semicircle has a diameter equal to the width W. So the circumference of a full circle with diameter W is πW, so two semicircles would be πW. Therefore, the total length of the track is 2L + πW = 400 meters.

So the problem says the length of the track is 400 meters. So that equation is 2L + πW = 400.

And we need to find the maximum area of the field. The area of the field is length times width, so A = L * W.

So we need to maximize A = L * W, given that 2L + πW = 400.

This is an optimization problem with a constraint. So we can use methods like substitution or Lagrange multipliers. But since it's a single constraint with two variables, substitution seems straightforward.

Let me solve the constraint equation for one variable in terms of the other. Let's solve for L.

2L = 400 - πW

So L = (400 - πW)/2

Then substitute into the area formula:

A = L * W = [(400 - πW)/2] * W = (400W - πW²)/2

So A = 200W - (π/2)W²

Now, to find the maximum area, we can take the derivative of A with respect to W and set it equal to zero.

But wait, since this is a quadratic function in terms of W, it's a downward-opening parabola, so the maximum occurs at the vertex.

Alternatively, I can use calculus. Let me try calculus here.

dA/dW = 200 - πW

Set derivative equal to zero:

200 - πW = 0 => πW = 200 => W = 200 / π

Then, substitute back into the equation for L:

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

So the maximum area occurs when W = 200/π and L = 100.

Then the area A is L * W = 100 * (200 / π) = 20000 / π

Wait, but let me check if this is correct. Alternatively, using the vertex formula for a quadratic.

The quadratic equation is A = - (π/2)W² + 200W.

In general, for a quadratic ax² + bx + c, the vertex is at x = -b/(2a). Here, a = -π/2, b = 200.

So W = -200 / (2*(-π/2)) = -200 / (-π) = 200/π. Yep, same result.

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

But let me check if the problem is referring to the area of the field, which is the rectangle. Wait, but the Asymptote code shows a rectangle with semicircular ends, but maybe the "field" refers to the rectangle? Or is the entire track area? Wait, the problem says "a rectangular field is enclosed in an athletic track". So the field is the rectangle, and the track is around it? Or is the track itself the field?

Wait, the problem says "the track consists of two edges of the field, and two semicircles." So the track is made up of two edges of the field (so the straight sides) and two semicircles. So maybe the field is the rectangle, and the track is surrounding it? Or is the track the field?

Wait, maybe I need to confirm. Let me re-read the problem.

"A rectangular field is enclosed in an athletic track, as shown below. The track consists of two edges of the field, and two semicircles. The length of the track is 400 meters. What is the largest possible area of the field, in square meters?"

So the rectangular field is enclosed by the track. The track consists of two edges of the field and two semicircles. So the track is made of the two straight sides of the field (the length of the rectangle) and two semicircular ends. Therefore, the total length of the track is the two lengths plus the circumference of the circle (since two semicircles make a full circle).

Therefore, as I had earlier, the total perimeter of the track is 2L + πW = 400.

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

So my previous calculation seems right.

But just to be thorough, let me check units. 2L + πW = 400 meters. So if W is in meters, then yes.

So maximum area is 20000 / π. Let me compute that numerically.

But maybe the question wants an exact value, so 20000/π. But maybe I can write it as (200^2)/π. Wait, 200*100 is 20000.

Alternatively, if I think about the formula, maybe there's a different approach.

Wait, another way to think about this is that the track is similar to a standard running track. In reality, the perimeter of a running track is calculated as 2*(length) + 2*(π*(diameter)/2), which is the same as 2L + πW. So that's consistent with our equation.

But perhaps there's a different way to think about maximizing area with a fixed perimeter? Wait, but in this case, the perimeter isn't exactly the same as a regular perimeter. The "track" has a different shape, but the problem defines its length as 400 meters.

Alternatively, perhaps if we think of the entire track as a closed loop. If the track is made up of two straight sides of length L and two semicircular ends with diameter W, then the total length of the track is 2L + πW. So that's given as 400.

Therefore, the problem reduces to maximizing A = L*W with 2L + πW = 400. Which is exactly what I did earlier.

Therefore, solving gives W = 200/π, L = 100, A = 20000/π.

But let me check if this is the maximum. For example, if we take W approaching zero, then L approaches 200, and area approaches zero. If W approaches 400/π, then L approaches zero, and area approaches zero. So the maximum is indeed somewhere in between.

Alternatively, if I think of this as an optimization problem, maybe using Lagrange multipliers. Let me try that.

Let me set up the function to maximize: A = L*W.

Subject to the constraint: 2L + πW = 400.

Using Lagrange multipliers, set up ∇A = λ∇g, where g(L, W) = 2L + πW - 400 = 0.

Compute the gradients:

∇A = (W, L)

∇g = (2, π)

So setting up equations:

W = 2λ

L = πλ

So from the first equation, λ = W/2. Substitute into second equation: L = π*(W/2) => L = (π/2)W.

But from the constraint equation: 2L + πW = 400. Substitute L = (π/2)W:

2*(π/2)W + πW = πW + πW = 2πW = 400 => W = 400/(2π) = 200/π. Then L = (π/2)*(200/π) = 100.

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

So, 20000 divided by pi.

But maybe I need to check if the problem is referring to the area of the track or the field? Wait, the problem says "the largest possible area of the field". The field is the rectangle, so that's L*W. So yes, 20000/pi.

But maybe I need to verify with some numbers. Let me take a concrete example.

Suppose that W = 200/pi ≈ 200/3.1416 ≈ 63.66 meters. Then L = 100 meters.

Then perimeter is 2*100 + π*63.66 ≈ 200 + 200 = 400 meters. Which checks out.

So the area would be 100*63.66 ≈ 6366 square meters.

Alternatively, if I take different values. Suppose W = 50 meters, then 2L + π*50 = 400 => 2L = 400 - 157.08 = 242.92 => L = 121.46. Then area is 121.46*50 ≈ 6073, which is less than 6366.

Another example: let me take W = 70 meters. Then 2L + π*70 ≈ 2L + 219.91 = 400 => 2L ≈ 180.09 => L ≈ 90.045. Area ≈ 90.045 * 70 ≈ 6303. Which is still less than 6366.

So seems like 20000/pi is the maximum.

Alternatively, if I take W = 200/pi ≈ 63.66 meters. Then the area is approx 20000/3.1416 ≈ 6366.2 square meters.

So I think that's correct.

But wait, the problem mentions that the track consists of two edges of the field and two semicircles. Is there a possibility that the semicircles are on the length instead of the width? Wait, no. If the track consists of two edges of the field and two semicircles, the semicircles have to be on the ends of the field. So if the field is a rectangle, the two semicircles would be on the shorter sides, so the diameter of the semicircles would be the width of the field.

Alternatively, if the semicircles were on the length, then the diameter would be length, but that would make the track longer. But given that the track length is 400 meters, which is a standard track length. For example, a standard 400m track has two straightaways and two semicircular ends. In reality, the straightaways are about 84.39 meters each, and the diameter of the semicircular ends is about 73 meters. But of course, the exact numbers depend on the design. However, in our problem, we are to maximize the area of the field (the rectangle).

So, in the calculations, we found that when the width is 200/pi and length is 100, the area is maximized.

Alternatively, if the problem was asking for the area of the entire track, which would be the area of the rectangle plus the area of the circle (since two semicircles make a full circle). But the problem says "the largest possible area of the field", so the field is the rectangle, so the answer should be 20000/pi.

But let me just check if I didn't misinterpret the track's perimeter.

The problem says "the track consists of two edges of the field, and two semicircles". So, if the field is the rectangle, then the track is built along two sides of the field (the length?) and the semicircular ends.

Wait, if the track is surrounding the field, then maybe the two edges of the field are the straight sides of the track. So if the field is a rectangle, then the track would have two straight sides equal to the length of the field and two semicircular ends with diameter equal to the width of the field. Therefore, the total length of the track is 2L + πW.

Alternatively, if the track was built on the field, meaning that the field is inside the track. But the problem says the field is enclosed in the track. So the field is inside. Therefore, the track is the outer perimeter.

Therefore, in that case, the length of the track would be the outer perimeter, which would be 2L + 2*(πW/2) = 2L + πW. Which is again the same equation.

Therefore, the calculation is correct.

Thus, the maximum area of the field is 20000 divided by pi.

But just to confirm, let's think about the standard maximum area for a given perimeter. For a given perimeter, the maximum area is achieved by a circle. However, in this case, the shape is constrained to be a rectangle with semicircular ends, so the maximum area under this constraint is different.

But in this case, we derived that the maximum area is 20000/pi.

Alternatively, if we think of the entire track as a shape with perimeter 400m, but the track has a different shape. However, the problem specifically says that the track consists of two edges of the field and two semicircles, so we have to stick with that.

Therefore, I think my answer is correct.

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

To solve the problem, we interpret the description as follows: a rectangular field is enclosed by a track. The track comprises **two straight sides** of the field (the sides of the rectangle) and **two semicircles** at the ends of the field. Since two semicircles make a full circle, the total length of the track is the sum of the **perimeter of the two straight sides** and the **circumference of a full circle**.

---

### Step 1: Define Variables
Let:
- $ L $ be the **length** of the rectangular field
- $ W $ be the **width** of the rectangular field

The straight segments of the track are the two lengths of the rectangle, so their total is:
$$
2L
$$

The curved part is a full circle with **diameter $ W $**, so its circumference is:
$$
\pi W
$$

Therefore, the total length of the track is:
$$
2L + \pi W = 400 \quad \text{(1)}
$$

The **area of the field**, which is the rectangle, is:
$$
A = L \cdot W \quad \text{(2)}
$$

Our goal is to **maximize** this area, $ A $, under the constraint given by equation (1).

---

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

From equation (1), solve for $ L $:
$$
2L = 400 - \pi W \Rightarrow L = \frac{400 - \pi W}{2}
$$

Substitute into equation (2):
$$
A = L \cdot W = \frac{400 - \pi W}{2} \cdot W = (200 - \frac{\pi}{2} W) \cdot W = 200W - \frac{\pi}{2} W^2
$$

---

### Step 3: Maximize the Area Function

We now analyze the quadratic function:
$$
A(W) = 200W - \frac{\pi}{2} W^2
$$

This is a **downward-opening parabola**, so it has a **maximum** at the vertex. The vertex occurs when:
$$
W = \frac{-b}{2a}
$$

In this case:
- $ a = -\frac{\pi}{2} $
- $ b = 200 $

So:
$$
W = \frac{-200}{2 \cdot (-\frac{\pi}{2})} = \frac{-200}{-\pi} = \frac{200}{\pi}
$$

---

### Step 4: Find the Maximum Area

Now substitute $ W = \frac{200}{\pi} $ back into the expression for $ A $:

First, find $ L $:
$$
L = \frac{400 - \pi W}{2} = \frac{400 - \pi \cdot \frac{200}{\pi}}{2} = \frac{400 - 200}{2} = 100
$$

Then compute the 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, in square meters.

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 me try to figure out how to solve this problem. So, Kathy has $300 in the bank, and she wants to withdraw half of it in British pounds and half in euros. The question is asking how many more euros than pounds she will have. They also gave the exchange rates: 1 pound equals 1.64 USD and 1 euro equals 1.32 USD. And we need to round to the nearest whole number. Alright, let me break this down step by step.

First, let me understand what's being asked. She wants to withdraw half of her $300 as pounds and the other half as euros. So, half of $300 is $150. That means she wants to convert $150 into pounds and $150 into euros. Then, we need to find out how many euros she gets from the $150 and how many pounds she gets from the other $150. Then, subtract the number of pounds from the number of euros to find out how many more euros she has than pounds. Got it.

Let me start with converting the $150 to pounds. The exchange rate is 1 pound = 1.64 USD. So, to find out how many pounds she gets for $150, I need to divide the amount in USD by the exchange rate. That is, pounds = USD / 1.64. So, pounds = 150 / 1.64. Let me calculate that.

Hmm, 150 divided by 1.64. Let me do this division. Maybe I can simplify this. Let me write it down:

150 ÷ 1.64

To make this easier, I can multiply numerator and denominator by 100 to eliminate the decimal. So, that becomes 15000 ÷ 164. Let me compute that.

Alternatively, maybe I can approximate. Let me think. 1.64 times 90 is 147.6, right? 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. So, approximately 91.46 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 would be 91.463. So, approximately 91.46 pounds. So, that's about 91.46 pounds. But let me confirm with actual division.

Let me try dividing 150 by 1.64. Let's do this more accurately. Let me write it as 150 ÷ 1.64. Since 1.64 goes into 150 how many times?

Alternatively, maybe convert 1.64 to a fraction? Not sure if that helps. Alternatively, maybe use decimal division.

Let me set it up as 1.64 | 150.00

First, since 1.64 goes into 150 how many times?

Multiply 1.64 by 90: as above, 147.6. Subtract that from 150: 150 - 147.6 = 2.4

Bring down a zero: 24.0

Now, 1.64 goes into 24.0 how many times? 1.64 * 14 = 22.96. That's close. So 14 times. 14*1.64 = 22.96. Subtract that from 24.0: 24.0 - 22.96 = 1.04

Bring down another zero: 10.40

1.64 goes into 10.40 how many times? 1.64*6 = 9.84. Subtract from 10.40: 10.40 - 9.84 = 0.56

Bring down another zero: 5.60

1.64 goes into 5.60 three times? 1.64*3=4.92. Subtract: 5.60 -4.92 = 0.68

Bring down another zero: 6.80

1.64*4=6.56. Subtract: 6.80 -6.56=0.24

Bring down another zero: 2.40

Wait, this is starting to repeat. So putting it all together, we had 90 + 14 + 6 + 3 + 4... Wait, no, actually, the initial division steps:

First, 1.64 into 150 is 91.46... So after the decimal, we had 91.4... So, the pounds are approximately 91.46 pounds. So, maybe 91.46 pounds. Let me check with calculator-like steps.

Alternatively, maybe I can use fractions. Let me think. 150 divided by 1.64. Let me write 1.64 as 164/100. Therefore, 150 divided by (164/100) is 150 * (100/164) = (150*100)/164 = 15000/164. Let me compute that.

Divide numerator and denominator by 4: 15000 ÷ 4 = 3750; 164 ÷4=41. So, 3750 / 41. Let me divide 3750 by 41. 41*90 = 3690. Subtract that from 3750: 3750 - 3690 = 60. Then, 41*1=41. Subtract: 60 -41=19. Bring down a zero: 190. 41*4=164. Subtract: 190 -164=26. Bring down a zero: 260. 41*6=246. Subtract: 260 -246=14. Bring down a zero: 140. 41*3=123. Subtract: 140 -123=17. So, so far, we have 90 +1 + 0.4 + 0.06 + 0.003... So, it's approximately 91.463... So, 91.46 pounds. So, rounding to the nearest whole number? Wait, no, hold on. Wait, the question says to round to the nearest whole number at the end. Wait, but maybe we need to keep more decimal places for accuracy?

Wait, but let me check. The problem says "how many more euros than pounds will she have?" and "round to the nearest whole number." So, we need to compute the exact values for euros and pounds and then subtract, then round the final result. So, maybe I need to keep more precise numbers for both euros and pounds?

Alternatively, maybe I can compute both amounts with more precision and then subtract.

But let's proceed step by step.

So, first, converting $150 to pounds: 150 / 1.64 ≈ 91.46341463... So approximately 91.46 pounds.

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

Again, let me compute that. Let me do the same process.

So, 150 ÷ 1.32. Let me convert that to fractions again. 1.32 is 132/100. So 150 divided by (132/100) is 150*(100/132) = 15000 / 132. Let me simplify that.

Divide numerator and denominator by 12: 15000 ÷12 = 1250; 132 ÷12 = 11. So 1250 / 11. Let me compute that division.

11*113 = 1243. Because 11*100=1100, 11*13=143; 1100+143=1243. Then, 1250 - 1243 = 7. So, 1250 / 11 = 113 + 7/11 ≈ 113.6363... So approximately 113.636 euros.

Alternatively, let me check with decimal division:

1.32 | 150.00

First, 1.32 goes into 150 how many times?

Estimate: 1.32 * 100 = 132. So subtract 132 from 150, we get 18. Bring down a zero, making it 180. 1.32 goes into 180 how many times? Let me calculate 1.32*136 = ?

Wait, maybe another approach. Let me think. 1.32 * 113 = ?

Well, 1.32 * 100 = 132

1.32 * 10 = 13.2

1.32 * 3 = 3.96

So total 1.32*(100 + 10 + 3) = 132 + 13.2 + 3.96 = 132 + 17.16 = 149.16

So 1.32 * 113 = 149.16. Then, 150.00 - 149.16 = 0.84. Bring down a zero: 8.40.

Now, 1.32 goes into 8.40 how many times? 1.32 * 6 = 7.92. Subtract from 8.40: 8.40 - 7.92 = 0.48. Bring down a zero: 4.80. 1.32 goes into 4.80 three times (1.32*3 = 3.96). Subtract: 4.80 - 3.96 = 0.84. Bring down a zero: 8.40... So it's repeating. So, so far, we have 113.63... So, approximately 113.636 euros.

So, euros ≈ 113.636 and pounds ≈ 91.463.

Now, the question is how many more euros than pounds. So, euros - pounds = 113.636 - 91.463 ≈ 22.173. So approximately 22.17 euros more. Now, we need to round to the nearest whole number. So 22.17 is approximately 22. So, the answer is 22?

Wait, but let me check if my calculations are correct. Because maybe I approximated too early. Maybe I should carry out the exact decimal divisions with more precision?

Alternatively, let me compute the exact values using fractions. Let me check again.

First, pounds: 150 / 1.64. Let me write 1.64 as 41/25? 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, 150 divided by (41/25) is 150 * (25/41) = (150*25)/41.

Similarly, euros: 150 divided by 1.32. 1.32 is 33/25? Wait, 1.32 is 132/100 = 33/25. Yes, because 132 divided by 4 is 33, 100 divided by 4 is 25. So 1.32 is 33/25. Therefore, 150 divided by (33/25) is 150*(25/33) = (150*25)/33.

So, let me compute these fractions.

First, pounds: (150 * 25)/41 = (3750)/41 ≈ ?

Let me compute 3750 divided by 41.

Well, 41*90 = 3690.

Subtract 3690 from 3750: 3750 - 3690 = 60.

So, 60/41 ≈ 1.4634...

So total is 90 + 1.4634 ≈ 91.4634 pounds.

Similarly, euros: (150*25)/33 = 3750 / 33 ≈ ?

Let me compute 33*113 = 33*(100 + 13) = 3300 + 429 = 3729.

Subtract 3729 from 3750: 3750 - 3729 = 21.

So 21/33 = 7/11 ≈ 0.6364...

Therefore, euros ≈ 113 + 0.6364 ≈ 113.6364 euros.

Therefore, euros - pounds ≈ 113.6364 - 91.4634 ≈ 22.173.

So, 22.173. Rounded to the nearest whole number is 22. Because 0.173 is less than 0.5, so we round down. So the answer should be 22.

Wait, but let me check if there was any miscalculations here. Let me verify the exact decimal values once again.

Alternatively, perhaps I can use calculator-like approximations for better accuracy?

Alternatively, maybe I should carry out the exact decimal computations with more precision?

Wait, let me check the euros again. 3750 divided by 33. Let me do this division again.

So 33 into 3750.

First, 33*100 = 3300. So 100 times. 3750 - 3300 = 450.

Now, 33*13 = 429. So subtract 429 from 450: 450 - 429 = 21. So total is 100 + 13 = 113, with a remainder of 21. So 21/33 = 0.6363..., so 113.6363...

Similarly, pounds was 91.4634...

So the difference is 113.6363... - 91.4634... Let me compute that.

Subtracting the whole numbers: 113 - 91 = 22. Then subtract the decimals: 0.6363... - 0.4634... = ?

Compute 0.6363 - 0.4634:

0.6363 - 0.4634 = 0.1729...

So total difference is 22 + 0.1729 = 22.1729... So approximately 22.1729. So, 22.17 when rounded to the nearest hundredth. Rounded to the nearest whole number is 22, since 0.17 is less than 0.5.

Therefore, the answer is 22.

But let me think again if there's another approach where perhaps the rounding is done at each step, which could affect the result? For example, if we round the pounds and euros to the nearest whole number first, before subtracting, would that change the answer?

Let me check. Suppose we round pounds to the nearest whole number: 91.463 is approximately 91.46, which is 91 when rounded to the nearest whole number (since 0.46 is less than 0.5). Similarly, euros is 113.636, which is approximately 114 when rounded to the nearest whole number (since 0.636 is more than 0.5). Then, 114 - 91 = 23. But the exact difference was 22.17, which rounds to 22. So, depending on when you round, you might get 22 or 23. However, the problem says "round to the nearest whole number" at the end, so we should carry out the exact calculation first and then round at the end. Therefore, 22.17 rounds to 22. Therefore, the answer is 22.

But let me confirm once more with another method. Maybe compute the values with more precise decimal places?

Alternatively, perhaps use decimal multiplication for the exchange rates?

Alternatively, let me check my initial calculation with another approach.

For pounds: $150 converted to pounds. Since 1 pound is 1.64 USD, then 1 USD is 1/1.64 pounds. So, 150 USD is 150*(1/1.64) pounds. Let me compute 1/1.64.

1 divided by 1.64. Let me calculate that. 1.64 goes into 1.00 how many times?

Well, 1/1.64 ≈ 0.609756... So multiplying by 150, 150*0.609756 ≈ ?

Let me compute that. 150 * 0.6 = 90. 150 * 0.009756 ≈ 150 * 0.01 = 1.5, so subtract 150*(0.01 - 0.009756) = 150*(0.000244) ≈ 0.0366. Therefore, approximately 1.5 - 0.0366 ≈ 1.4634. Therefore, total pounds ≈ 90 + 1.4634 ≈ 91.4634. Same as before.

Similarly for euros: 1 euro is 1.32 USD, so 1 USD is 1/1.32 euros. Let me compute 1/1.32. 1 divided by 1.32. Let me calculate that. 1.32 goes into 1.00 how many times? Approximately 0.757575... because 1/1.32 = 100/132 = 25/33 ≈ 0.757575...

Therefore, 150 USD * 25/33 euros per USD = (150*25)/33 = 3750/33 ≈ 113.6363... Same as before. Therefore, difference is 113.6363 - 91.4634 ≈ 22.1729.

So, 22.1729, which is approximately 22.17, rounds to 22. Therefore, answer is 22.

But let me check if I have misread the question. The question says "how many more euros than pounds will she have?" So, euros minus pounds. So, yes, 113.636 - 91.463 ≈ 22.17. So, 22 when rounded.

Alternatively, maybe the question wants the difference in USD? But no, the question says "how many more euros than pounds will she have?" So, the answer should be in euros, but since we are comparing euros and pounds, the difference is in terms of the number of euros or pounds? Wait, no. Let me think.

Wait, when they say "how many more euros than pounds", they mean if you take the amount in euros and subtract the amount in pounds, what is the result? But since euros and pounds are different currencies, the question is a bit ambiguous. However, given that they provide exchange rates and ask to round to the nearest whole number, it's likely that they expect the numerical difference after converting both amounts to their respective currencies, and then subtracting. So, for example, if she has X euros and Y pounds, how much more euros than pounds? That would be X - Y, where X and Y are in their respective units. So, the answer is approximately 22.17, which rounds to 22. So, 22 euros more than pounds.

Alternatively, maybe they want to convert both to USD and then subtract? Wait, but that would not make sense because the question says "how many more euros than pounds", not "how many more dollars".

But let me think. Suppose she has some amount in euros and some in pounds. If someone asks how many more euros than pounds she has, they would mean the numerical difference in the number of euros and the number of pounds. So, if she has 113 euros and 91 pounds, she has 22 more euros than pounds. So, the answer is 22. Therefore, my answer is 22. So, I think that is correct.

But just to make sure, let me check if there's any possible mistake in exchange rates.

Given that 1 pound = 1.64 USD and 1 euro = 1.32 USD. So, to convert USD to pounds, divide by 1.64; to convert USD to euros, divide by 1.32. Which is what I did. So, 150 USD is 150 / 1.64 pounds and 150 / 1.32 euros. So that is correct.

Alternatively, if someone thought that you need to convert the pounds and euros back to USD, but that would be unnecessary here.

Alternatively, maybe the question is trying to trick us into thinking that the amounts are in USD, but no. The problem says she wants to withdraw half in pounds and half in euros. So, the total amount in USD is 300, so half is 150 USD. So, converting that 150 USD into pounds and 150 USD into euros. So, calculations are as above.

Another check: Let me compute the total value in USD after conversion. For instance, if she has X pounds and Y euros, then total USD would be X*1.64 + Y*1.32. But since she took out 150 USD for pounds and 150 USD for euros, the total value should be 150 + 150 = 300 USD. Let me check if that's the case.

Compute X*1.64 + Y*1.32 where X ≈ 91.463 and Y ≈ 113.636.

Compute 91.463 * 1.64 + 113.636 * 1.32.

Let me compute 91.463 * 1.64:

First, 90 * 1.64 = 147.6

1.463 * 1.64 ≈ ?

Well, 1 * 1.64 = 1.64

0.463 * 1.64 ≈ Let me compute 0.4 * 1.64 = 0.656, 0.063 * 1.64 ≈ 0.10332. So total ≈ 0.656 + 0.10332 ≈ 0.75932. Therefore, total ≈ 1.64 + 0.75932 ≈ 2.39932. Therefore, total 91.463 * 1.64 ≈ 147.6 + 2.399 ≈ 149.999 ≈ 150 USD.

Similarly, 113.636 * 1.32:

Let me compute 113.636 * 1.32. Let's break it down:

First, 100 * 1.32 = 132

13.636 * 1.32. Let me compute 13 * 1.32 = 17.16, 0.636 * 1.32 ≈ 0.838. So total ≈ 17.16 + 0.838 ≈ 17.998. Therefore total ≈ 132 + 17.998 ≈ 149.998 ≈ 150 USD. Therefore, total is approximately 150 + 150 = 300 USD. That checks out. So, that confirms that the conversion is correct. Therefore, our earlier calculation of euros - pounds ≈ 22.17 is accurate.

Therefore, rounding to the nearest whole number is 22. Therefore, the answer is 22.

But just to ensure, let me try computing the exact difference without rounding.

We had euros ≈ 113.636363... and pounds ≈ 91.463414... The exact decimal difference is 113.636363... - 91.463414... Let me compute that.

First, write them as decimals:

Euros: 113.636363636...

Pounds: 91.463414634...

Subtract:

113.636363636...

-91.463414634...

= ?

Let me subtract digit by digit.

Starting from the decimal:

The integer parts: 113 - 91 = 22.

Decimal parts: 0.636363636... - 0.463414634...

Compute 0.636363636... - 0.463414634...

First, 0.636363... - 0.463414... Let me compute:

0.636363... = 0.636363...

0.463414... = 0.463414...

Subtract:

Start with 0.636363...

-0.463414...

Let me align decimals:

0.636363636...

-0.463414634...

= ?

Subtract:

Starting from the first decimal place:

0.6 - 0.4 = 0.2, but need to consider the following decimals.

But this might get complex. Alternatively, note that the decimal difference is approximately 0.1729... as before, leading to total difference of 22.1729...

Which is approximately 22.17, as before. So, 22.17 rounds to 22.

Therefore, after multiple checks and methods, I can confirm that the answer is 22.

**Final Answer**
\boxed{22}

To determine how many more euros than pounds Kathy will have, we start with the amount she wishes to convert: **$300**, with **half** of it ($150) converted to **British pounds (GBP)**, and the other **half** ($150) converted to **Euros (EUR)**.

---

### **Step 1: Convert USD to Pounds**

We are given the exchange rate:

$$
1 \text{ pound} = 1.64 \text{ USD}
$$

So, to find how many pounds $150 buys:

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

---

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

We are given the exchange rate:

$$
1 \text{ euro} = 1.32 \text{ USD}
$$

So, to find how many euros $150 buys:

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

---

### **Step 3: Calculate Difference Between Euros and Pounds**

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

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

---

### **Step 4: Round to the Nearest Whole Number**

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

---

### **Final Answer**

$$
\boxed{22}
$$

[Assistant]
Explanation:
Okay, let me try to figure out how to solve this problem. So, Kathy has $300 in the bank, and she wants to withdraw half of it in British pounds and half in euros. The question is asking how many more euros than pounds she will have. They also gave the exchange rates: 1 pound equals 1.64 USD and 1 euro equals 1.32 USD. And we need to round to the nearest whole number. Alright, let me break this down step by step.

First, let me understand what's being asked. She wants to withdraw half of her $300 as pounds and the other half as euros. So, half of $300 is $150. That means she wants to convert $150 into pounds and $150 into euros. Then, we need to find out how many euros she gets from the $150 and how many pounds she gets from the other $150. Then, subtract the number of pounds from the number of euros to find out how many more euros she has than pounds. Got it.

Let me start with converting the $150 to pounds. The exchange rate is 1 pound = 1.64 USD. So, to find out how many pounds she gets for $150, I need to divide the amount in USD by the exchange rate. That is, pounds = USD / 1.64. So, pounds = 150 / 1.64. Let me calculate that.

Hmm, 150 divided by 1.64. Let me do this division. Maybe I can simplify this. Let me write it down:

150 ÷ 1.64

To make this easier, I can multiply numerator and denominator by 100 to eliminate the decimal. So, that becomes 15000 ÷ 164. Let me compute that.

Alternatively, maybe I can approximate. Let me think. 1.64 times 90 is 147.6, right? 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. So, approximately 91.46 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 would be 91.463. So, approximately 91.46 pounds. So, that's about 91.46 pounds. But let me confirm with actual division.

Let me try dividing 150 by 1.64. Let's do this more accurately. Let me write it as 150 ÷ 1.64. Since 1.64 goes into 150 how many times?

Alternatively, maybe convert 1.64 to a fraction? Not sure if that helps. Alternatively, maybe use decimal division.

Let me set it up as 1.64 | 150.00

First, since 1.64 goes into 150 how many times?

Multiply 1.64 by 90: as above, 147.6. Subtract that from 150: 150 - 147.6 = 2.4

Bring down a zero: 24.0

Now, 1.64 goes into 24.0 how many times? 1.64 * 14 = 22.96. That's close. So 14 times. 14*1.64 = 22.96. Subtract that from 24.0: 24.0 - 22.96 = 1.04

Bring down another zero: 10.40

1.64 goes into 10.40 how many times? 1.64*6 = 9.84. Subtract from 10.40: 10.40 - 9.84 = 0.56

Bring down another zero: 5.60

1.64 goes into 5.60 three times? 1.64*3=4.92. Subtract: 5.60 -4.92 = 0.68

Bring down another zero: 6.80

1.64*4=6.56. Subtract: 6.80 -6.56=0.24

Bring down another zero: 2.40

Wait, this is starting to repeat. So putting it all together, we had 90 + 14 + 6 + 3 + 4... Wait, no, actually, the initial division steps:

First, 1.64 into 150 is 91.46... So after the decimal, we had 91.4... So, the pounds are approximately 91.46 pounds. So, maybe 91.46 pounds. Let me check with calculator-like steps.

Alternatively, maybe I can use fractions. Let me think. 150 divided by 1.64. Let me write 1.64 as 164/100. Therefore, 150 divided by (164/100) is 150 * (100/164) = (150*100)/164 = 15000/164. Let me compute that.

Divide numerator and denominator by 4: 15000 ÷ 4 = 3750; 164 ÷4=41. So, 3750 / 41. Let me divide 3750 by 41. 41*90 = 3690. Subtract that from 3750: 3750 - 3690 = 60. Then, 41*1=41. Subtract: 60 -41=19. Bring down a zero: 190. 41*4=164. Subtract: 190 -164=26. Bring down a zero: 260. 41*6=246. Subtract: 260 -246=14. Bring down a zero: 140. 41*3=123. Subtract: 140 -123=17. So, so far, we have 90 +1 + 0.4 + 0.06 + 0.003... So, it's approximately 91.463... So, 91.46 pounds. So, rounding to the nearest whole number? Wait, no, hold on. Wait, the question says to round to the nearest whole number at the end. Wait, but maybe we need to keep more decimal places for accuracy?

Wait, but let me check. The problem says "how many more euros than pounds will she have?" and "round to the nearest whole number." So, we need to compute the exact values for euros and pounds and then subtract, then round the final result. So, maybe I need to keep more precise numbers for both euros and pounds?

Alternatively, maybe I can compute both amounts with more precision and then subtract.

But let's proceed step by step.

So, first, converting $150 to pounds: 150 / 1.64 ≈ 91.46341463... So approximately 91.46 pounds.

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

Again, let me compute that. Let me do the same process.

So, 150 ÷ 1.32. Let me convert that to fractions again. 1.32 is 132/100. So 150 divided by (132/100) is 150*(100/132) = 15000 / 132. Let me simplify that.

Divide numerator and denominator by 12: 15000 ÷12 = 1250; 132 ÷12 = 11. So 1250 / 11. Let me compute that division.

11*113 = 1243. Because 11*100=1100, 11*13=143; 1100+143=1243. Then, 1250 - 1243 = 7. So, 1250 / 11 = 113 + 7/11 ≈ 113.6363... So approximately 113.636 euros.

Alternatively, let me check with decimal division:

1.32 | 150.00

First, 1.32 goes into 150 how many times?

Estimate: 1.32 * 100 = 132. So subtract 132 from 150, we get 18. Bring down a zero, making it 180. 1.32 goes into 180 how many times? Let me calculate 1.32*136 = ?

Wait, maybe another approach. Let me think. 1.32 * 113 = ?

Well, 1.32 * 100 = 132

1.32 * 10 = 13.2

1.32 * 3 = 3.96

So total 1.32*(100 + 10 + 3) = 132 + 13.2 + 3.96 = 132 + 17.16 = 149.16

So 1.32 * 113 = 149.16. Then, 150.00 - 149.16 = 0.84. Bring down a zero: 8.40.

Now, 1.32 goes into 8.40 how many times? 1.32 * 6 = 7.92. Subtract from 8.40: 8.40 - 7.92 = 0.48. Bring down a zero: 4.80. 1.32 goes into 4.80 three times (1.32*3 = 3.96). Subtract: 4.80 - 3.96 = 0.84. Bring down a zero: 8.40... So it's repeating. So, so far, we have 113.63... So, approximately 113.636 euros.

So, euros ≈ 113.636 and pounds ≈ 91.463.

Now, the question is how many more euros than pounds. So, euros - pounds = 113.636 - 91.463 ≈ 22.173. So approximately 22.17 euros more. Now, we need to round to the nearest whole number. So 22.17 is approximately 22. So, the answer is 22?

Wait, but let me check if my calculations are correct. Because maybe I approximated too early. Maybe I should carry out the exact decimal divisions with more precision?

Alternatively, let me compute the exact values using fractions. Let me check again.

First, pounds: 150 / 1.64. Let me write 1.64 as 41/25? 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, 150 divided by (41/25) is 150 * (25/41) = (150*25)/41.

Similarly, euros: 150 divided by 1.32. 1.32 is 33/25? Wait, 1.32 is 132/100 = 33/25. Yes, because 132 divided by 4 is 33, 100 divided by 4 is 25. So 1.32 is 33/25. Therefore, 150 divided by (33/25) is 150*(25/33) = (150*25)/33.

So, let me compute these fractions.

First, pounds: (150 * 25)/41 = (3750)/41 ≈ ?

Let me compute 3750 divided by 41.

Well, 41*90 = 3690.

Subtract 3690 from 3750: 3750 - 3690 = 60.

So, 60/41 ≈ 1.4634...

So total is 90 + 1.4634 ≈ 91.4634 pounds.

Similarly, euros: (150*25)/33 = 3750 / 33 ≈ ?

Let me compute 33*113 = 33*(100 + 13) = 3300 + 429 = 3729.

Subtract 3729 from 3750: 3750 - 3729 = 21.

So 21/33 = 7/11 ≈ 0.6364...

Therefore, euros ≈ 113 + 0.6364 ≈ 113.6364 euros.

Therefore, euros - pounds ≈ 113.6364 - 91.4634 ≈ 22.173.

So, 22.173. Rounded to the nearest whole number is 22. Because 0.173 is less than 0.5, so we round down. So the answer should be 22.

Wait, but let me check if there was any miscalculations here. Let me verify the exact decimal values once again.

Alternatively, perhaps I can use calculator-like approximations for better accuracy?

Alternatively, maybe I should carry out the exact decimal computations with more precision?

Wait, let me check the euros again. 3750 divided by 33. Let me do this division again.

So 33 into 3750.

First, 33*100 = 3300. So 100 times. 3750 - 3300 = 450.

Now, 33*13 = 429. So subtract 429 from 450: 450 - 429 = 21. So total is 100 + 13 = 113, with a remainder of 21. So 21/33 = 0.6363..., so 113.6363...

Similarly, pounds was 91.4634...

So the difference is 113.6363... - 91.4634... Let me compute that.

Subtracting the whole numbers: 113 - 91 = 22. Then subtract the decimals: 0.6363... - 0.4634... = ?

Compute 0.6363 - 0.4634:

0.6363 - 0.4634 = 0.1729...

So total difference is 22 + 0.1729 = 22.1729... So approximately 22.1729. So, 22.17 when rounded to the nearest hundredth. Rounded to the nearest whole number is 22, since 0.17 is less than 0.5.

Therefore, the answer is 22.

But let me think again if there's another approach where perhaps the rounding is done at each step, which could affect the result? For example, if we round the pounds and euros to the nearest whole number first, before subtracting, would that change the answer?

Let me check. Suppose we round pounds to the nearest whole number: 91.463 is approximately 91.46, which is 91 when rounded to the nearest whole number (since 0.46 is less than 0.5). Similarly, euros is 113.636, which is approximately 114 when rounded to the nearest whole number (since 0.636 is more than 0.5). Then, 114 - 91 = 23. But the exact difference was 22.17, which rounds to 22. So, depending on when you round, you might get 22 or 23. However, the problem says "round to the nearest whole number" at the end, so we should carry out the exact calculation first and then round at the end. Therefore, 22.17 rounds to 22. Therefore, the answer is 22.

But let me confirm once more with another method. Maybe compute the values with more precise decimal places?

Alternatively, perhaps use decimal multiplication for the exchange rates?

Alternatively, let me check my initial calculation with another approach.

For pounds: $150 converted to pounds. Since 1 pound is 1.64 USD, then 1 USD is 1/1.64 pounds. So, 150 USD is 150*(1/1.64) pounds. Let me compute 1/1.64.

1 divided by 1.64. Let me calculate that. 1.64 goes into 1.00 how many times?

Well, 1/1.64 ≈ 0.609756... So multiplying by 150, 150*0.609756 ≈ ?

Let me compute that. 150 * 0.6 = 90. 150 * 0.009756 ≈ 150 * 0.01 = 1.5, so subtract 150*(0.01 - 0.009756) = 150*(0.000244) ≈ 0.0366. Therefore, approximately 1.5 - 0.0366 ≈ 1.4634. Therefore, total pounds ≈ 90 + 1.4634 ≈ 91.4634. Same as before.

Similarly for euros: 1 euro is 1.32 USD, so 1 USD is 1/1.32 euros. Let me compute 1/1.32. 1 divided by 1.32. Let me calculate that. 1.32 goes into 1.00 how many times? Approximately 0.757575... because 1/1.32 = 100/132 = 25/33 ≈ 0.757575...

Therefore, 150 USD * 25/33 euros per USD = (150*25)/33 = 3750/33 ≈ 113.6363... Same as before. Therefore, difference is 113.6363 - 91.4634 ≈ 22.1729.

So, 22.1729, which is approximately 22.17, rounds to 22. Therefore, answer is 22.

But let me check if I have misread the question. The question says "how many more euros than pounds will she have?" So, euros minus pounds. So, yes, 113.636 - 91.463 ≈ 22.17. So, 22 when rounded.

Alternatively, maybe the question wants the difference in USD? But no, the question says "how many more euros than pounds will she have?" So, the answer should be in euros, but since we are comparing euros and pounds, the difference is in terms of the number of euros or pounds? Wait, no. Let me think.

Wait, when they say "how many more euros than pounds", they mean if you take the amount in euros and subtract the amount in pounds, what is the result? But since euros and pounds are different currencies, the question is a bit ambiguous. However, given that they provide exchange rates and ask to round to the nearest whole number, it's likely that they expect the numerical difference after converting both amounts to their respective currencies, and then subtracting. So, for example, if she has X euros and Y pounds, how much more euros than pounds? That would be X - Y, where X and Y are in their respective units. So, the answer is approximately 22.17, which rounds to 22. So, 22 euros more than pounds.

Alternatively, maybe they want to convert both to USD and then subtract? Wait, but that would not make sense because the question says "how many more euros than pounds", not "how many more dollars".

But let me think. Suppose she has some amount in euros and some in pounds. If someone asks how many more euros than pounds she has, they would mean the numerical difference in the number of euros and the number of pounds. So, if she has 113 euros and 91 pounds, she has 22 more euros than pounds. So, the answer is 22. Therefore, my answer is 22. So, I think that is correct.

But just to make sure, let me check if there's any possible mistake in exchange rates.

Given that 1 pound = 1.64 USD and 1 euro = 1.32 USD. So, to convert USD to pounds, divide by 1.64; to convert USD to euros, divide by 1.32. Which is what I did. So, 150 USD is 150 / 1.64 pounds and 150 / 1.32 euros. So that is correct.

Alternatively, if someone thought that you need to convert the pounds and euros back to USD, but that would be unnecessary here.

Alternatively, maybe the question is trying to trick us into thinking that the amounts are in USD, but no. The problem says she wants to withdraw half in pounds and half in euros. So, the total amount in USD is 300, so half is 150 USD. So, converting that 150 USD into pounds and 150 USD into euros. So, calculations are as above.

Another check: Let me compute the total value in USD after conversion. For instance, if she has X pounds and Y euros, then total USD would be X*1.64 + Y*1.32. But since she took out 150 USD for pounds and 150 USD for euros, the total value should be 150 + 150 = 300 USD. Let me check if that's the case.

Compute X*1.64 + Y*1.32 where X ≈ 91.463 and Y ≈ 113.636.

Compute 91.463 * 1.64 + 113.636 * 1.32.

Let me compute 91.463 * 1.64:

First, 90 * 1.64 = 147.6

1.463 * 1.64 ≈ ?

Well, 1 * 1.64 = 1.64

0.463 * 1.64 ≈ Let me compute 0.4 * 1.64 = 0.656, 0.063 * 1.64 ≈ 0.10332. So total ≈ 0.656 + 0.10332 ≈ 0.75932. Therefore, total ≈ 1.64 + 0.75932 ≈ 2.39932. Therefore, total 91.463 * 1.64 ≈ 147.6 + 2.399 ≈ 149.999 ≈ 150 USD.

Similarly, 113.636 * 1.32:

Let me compute 113.636 * 1.32. Let's break it down:

First, 100 * 1.32 = 132

13.636 * 1.32. Let me compute 13 * 1.32 = 17.16, 0.636 * 1.32 ≈ 0.838. So total ≈ 17.16 + 0.838 ≈ 17.998. Therefore total ≈ 132 + 17.998 ≈ 149.998 ≈ 150 USD. Therefore, total is approximately 150 + 150 = 300 USD. That checks out. So, that confirms that the conversion is correct. Therefore, our earlier calculation of euros - pounds ≈ 22.17 is accurate.

Therefore, rounding to the nearest whole number is 22. Therefore, the answer is 22.

But just to ensure, let me try computing the exact difference without rounding.

We had euros ≈ 113.636363... and pounds ≈ 91.463414... The exact decimal difference is 113.636363... - 91.463414... Let me compute that.

First, write them as decimals:

Euros: 113.636363636...

Pounds: 91.463414634...

Subtract:

113.636363636...

-91.463414634...

= ?

Let me subtract digit by digit.

Starting from the decimal:

The integer parts: 113 - 91 = 22.

Decimal parts: 0.636363636... - 0.463414634...

Compute 0.636363636... - 0.463414634...

First, 0.636363... - 0.463414... Let me compute:

0.636363... = 0.636363...

0.463414... = 0.463414...

Subtract:

Start with 0.636363...

-0.463414...

Let me align decimals:

0.636363636...

-0.463414634...

= ?

Subtract:

Starting from the first decimal place:

0.6 - 0.4 = 0.2, but need to consider the following decimals.

But this might get complex. Alternatively, note that the decimal difference is approximately 0.1729... as before, leading to total difference of 22.1729...

Which is approximately 22.17, as before. So, 22.17 rounds to 22.

Therefore, after multiple checks and methods, I can confirm that the answer is 22.

**Final Answer**
\boxed{22}

To determine how many more euros than pounds Kathy will have, we start with the amount she wishes to convert: **$300**, with **half** of it ($150) converted to **British pounds (GBP)**, and the other **half** ($150) converted to **Euros (EUR)**.

---

### **Step 1: Convert USD to Pounds**

We are given the exchange rate:

$$
1 \text{ pound} = 1.64 \text{ USD}
$$

So, to find how many pounds $150 buys:

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

---

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

We are given the exchange rate:

$$
1 \text{ euro} = 1.32 \text{ USD}
$$

So, to find how many euros $150 buys:

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

---

### **Step 3: Calculate Difference Between Euros and Pounds**

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

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

---

### **Step 4: 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 start by recalling that the domain of a function involving a square root is all real numbers for which the expression inside the square root is non-negative. That means I need to figure out for which x values the expression (x - 3)^2 - (x - 8)^2 is greater than or equal to zero.

First, maybe I should simplify the expression inside the square root. Let me expand both squares. Let me compute (x - 3)^2 first. That would be x^2 - 6x + 9. Then, (x - 8)^2 is x^2 - 16x + 64. So if I subtract the second one from the first, it would be:

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

Let me distribute the negative sign to each term in the second polynomial:

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

Now combine like terms. The x^2 and -x^2 cancel each other out. Then the x terms: -6x + 16x is 10x. The constants: 9 - 64 is -55. So overall, the expression simplifies to 10x - 55.

So the original function simplifies to sqrt(10x - 55). Therefore, the domain of g(x) is all real numbers x such that 10x - 55 ≥ 0.

Solving that inequality: 10x - 55 ≥ 0 → 10x ≥ 55 → x ≥ 55/10 → x ≥ 11/2, which is 5.5.

Wait, so does that mean the domain is x ≥ 5.5? Then the smallest real number in the domain is 5.5? So the answer would be 11/2 or 5.5? Let me check my steps again to make sure I didn't make a mistake.

Starting with (x - 3)^2 - (x - 8)^2. Let me try expanding again to verify. Alternatively, maybe there's a formula for a^2 - b^2? Which is (a - b)(a + b). Maybe that could be a quicker way? Let me try that.

Let me let a = (x - 3) and b = (x - 8). Then the expression is a^2 - b^2 = (a - b)(a + b). Let's compute a - b and a + b.

a - b = (x - 3) - (x - 8) = x - 3 - x + 8 = 5.

a + b = (x - 3) + (x - 8) = 2x - 11.

Therefore, the expression (x - 3)^2 - (x - 8)^2 factors into 5*(2x - 11). Therefore, the expression inside the square root is 5*(2x - 11). So then, we need 5*(2x - 11) ≥ 0. Since 5 is positive, this is equivalent to 2x - 11 ≥ 0 → 2x ≥ 11 → x ≥ 11/2, which is 5.5. So same result as before. So that's good. So the domain is x ≥ 11/2, so the smallest real number in the domain is 11/2, which is 5.5.

But let me check once more with actual numbers. Let me plug in x = 5.5 into the original expression. Let me compute (5.5 - 3)^2 - (5.5 - 8)^2.

First, 5.5 - 3 is 2.5, so squared is 6.25. Then 5.5 - 8 is -2.5, squared is also 6.25. Therefore, 6.25 - 6.25 = 0. So sqrt(0) is 0, which is valid.

If I plug in x = 5.4, which is less than 5.5, then let's compute. 5.4 - 3 = 2.4, squared is 5.76. 5.4 - 8 = -2.6, squared is 6.76. Then 5.76 - 6.76 = -1.0, which is negative. Therefore, sqrt(-1.0) is not real, so x = 5.4 is not in the domain.

So indeed, x must be at least 5.5. Therefore, the smallest real number in the domain is 11/2.

But let me think again if there's another way this could be approached. Maybe I can consider the original expression under the square root. Let me see:

(x - 3)^2 - (x - 8)^2. Alternatively, maybe I can think of this as a difference of squares, but I already did that. Alternatively, think of it as quadratic in x. But since we simplified it to 10x - 55, which is linear, so the expression inside is linear. So the inequality is straightforward.

Alternatively, maybe there's some mistake in my algebra? Let me check the expansion again. First method:

(x - 3)^2 = x² - 6x + 9.

(x - 8)^2 = x² - 16x + 64.

Subtracting: (x² - 6x + 9) - (x² -16x + 64) = x² -6x +9 -x² +16x -64 = (x² -x²) + (-6x +16x) + (9 -64) = 0 +10x -55 = 10x -55. Correct.

Second method using difference of squares:

a^2 - b^2 = (a - b)(a + b). a = x -3, b = x -8. So, a - b = (x -3) - (x -8) = 5, as before. a + b = 2x -11. Therefore, 5*(2x -11). Which is 10x -55. Same result.

So both methods agree. So the expression inside the square root is 10x -55. Therefore, 10x -55 ≥ 0 → x ≥ 11/2. So the domain is [11/2, ∞). Therefore, the smallest real number in the domain is 11/2.

But just to make sure, maybe the problem is designed to trick us? Like maybe there's some hidden constraints? For example, sometimes with square roots, there might be more conditions, but in this case, the expression inside the square root simplifies to a linear function. So once we have that, the only constraint is 10x -55 ≥ 0.

Alternatively, could there be cases where even though 10x -55 is non-negative, there's some other restriction? For example, if there was a denominator or something else, but here the function is just sqrt(...). So no, there's no other restrictions. So the domain is indeed x ≥ 11/2.

So the smallest real number x in the domain is 11/2. Therefore, the answer is 11/2, which is 5.5, and as a fraction, 11/2.

But let me check if maybe the original problem had some typos or something? The problem says "smallest real number x in the domain", so if the domain is [11/2, ∞), then the smallest number is 11/2.

Alternatively, if the problem had something else, but no. So I think that is correct. Therefore, I think the answer is 11/2.

**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 begin by analyzing the expression inside the square root. A square root is defined only when its argument is non-negative (i.e., $ \geq 0 $). Therefore, the domain of $ g(x) $ consists of all real numbers $ x $ for which

$$
(x - 3)^2 - (x - 8)^2 \geq 0.
$$

---

### Step 1: Simplify the expression inside the square root

We simplify the expression step by step using algebra or a difference of squares identity:

$$
(x - 3)^2 - (x - 8)^2 = (x - 3 - (x - 8))(x - 3 + (x - 8)).
$$

Compute each factor:

$$
x - 3 - x + 8 = 5,\quad x - 3 + x - 8 = 2x - 11.
$$

So the expression simplifies to:

$$
(x - 3)^2 - (x - 8)^2 = 5(2x - 11) = 10x - 55.
$$

---

### Step 2: Determine when the expression is non-negative

We now need to find the values of $ x $ such that:

$$
10x - 55 \geq 0.
$$

Solving the inequality:

$$
10x \geq 55 \Rightarrow x \geq \frac{55}{10} = \frac{11}{2}.
$$

---

### Step 3: Interpret the result

This tells us that the domain of $ g(x) $ is:

$$
\left[\frac{11}{2}, \infty\right).
$$

Hence, the smallest real number in this domain is:

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

[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 start by recalling that the domain of a function involving a square root is all real numbers for which the expression inside the square root is non-negative. That means I need to figure out for which x values the expression (x - 3)^2 - (x - 8)^2 is greater than or equal to zero.

First, maybe I should simplify the expression inside the square root. Let me expand both squares. Let me compute (x - 3)^2 first. That would be x^2 - 6x + 9. Then, (x - 8)^2 is x^2 - 16x + 64. So if I subtract the second one from the first, it would be:

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

Let me distribute the negative sign to each term in the second polynomial:

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

Now combine like terms. The x^2 and -x^2 cancel each other out. Then the x terms: -6x + 16x is 10x. The constants: 9 - 64 is -55. So overall, the expression simplifies to 10x - 55.

So the original function simplifies to sqrt(10x - 55). Therefore, the domain of g(x) is all real numbers x such that 10x - 55 ≥ 0.

Solving that inequality: 10x - 55 ≥ 0 → 10x ≥ 55 → x ≥ 55/10 → x ≥ 11/2, which is 5.5.

Wait, so does that mean the domain is x ≥ 5.5? Then the smallest real number in the domain is 5.5? So the answer would be 11/2 or 5.5? Let me check my steps again to make sure I didn't make a mistake.

Starting with (x - 3)^2 - (x - 8)^2. Let me try expanding again to verify. Alternatively, maybe there's a formula for a^2 - b^2? Which is (a - b)(a + b). Maybe that could be a quicker way? Let me try that.

Let me let a = (x - 3) and b = (x - 8). Then the expression is a^2 - b^2 = (a - b)(a + b). Let's compute a - b and a + b.

a - b = (x - 3) - (x - 8) = x - 3 - x + 8 = 5.

a + b = (x - 3) + (x - 8) = 2x - 11.

Therefore, the expression (x - 3)^2 - (x - 8)^2 factors into 5*(2x - 11). Therefore, the expression inside the square root is 5*(2x - 11). So then, we need 5*(2x - 11) ≥ 0. Since 5 is positive, this is equivalent to 2x - 11 ≥ 0 → 2x ≥ 11 → x ≥ 11/2, which is 5.5. So same result as before. So that's good. So the domain is x ≥ 11/2, so the smallest real number in the domain is 11/2, which is 5.5.

But let me check once more with actual numbers. Let me plug in x = 5.5 into the original expression. Let me compute (5.5 - 3)^2 - (5.5 - 8)^2.

First, 5.5 - 3 is 2.5, so squared is 6.25. Then 5.5 - 8 is -2.5, squared is also 6.25. Therefore, 6.25 - 6.25 = 0. So sqrt(0) is 0, which is valid.

If I plug in x = 5.4, which is less than 5.5, then let's compute. 5.4 - 3 = 2.4, squared is 5.76. 5.4 - 8 = -2.6, squared is 6.76. Then 5.76 - 6.76 = -1.0, which is negative. Therefore, sqrt(-1.0) is not real, so x = 5.4 is not in the domain.

So indeed, x must be at least 5.5. Therefore, the smallest real number in the domain is 11/2.

But let me think again if there's another way this could be approached. Maybe I can consider the original expression under the square root. Let me see:

(x - 3)^2 - (x - 8)^2. Alternatively, maybe I can think of this as a difference of squares, but I already did that. Alternatively, think of it as quadratic in x. But since we simplified it to 10x - 55, which is linear, so the expression inside is linear. So the inequality is straightforward.

Alternatively, maybe there's some mistake in my algebra? Let me check the expansion again. First method:

(x - 3)^2 = x² - 6x + 9.

(x - 8)^2 = x² - 16x + 64.

Subtracting: (x² - 6x + 9) - (x² -16x + 64) = x² -6x +9 -x² +16x -64 = (x² -x²) + (-6x +16x) + (9 -64) = 0 +10x -55 = 10x -55. Correct.

Second method using difference of squares:

a^2 - b^2 = (a - b)(a + b). a = x -3, b = x -8. So, a - b = (x -3) - (x -8) = 5, as before. a + b = 2x -11. Therefore, 5*(2x -11). Which is 10x -55. Same result.

So both methods agree. So the expression inside the square root is 10x -55. Therefore, 10x -55 ≥ 0 → x ≥ 11/2. So the domain is [11/2, ∞). Therefore, the smallest real number in the domain is 11/2.

But just to make sure, maybe the problem is designed to trick us? Like maybe there's some hidden constraints? For example, sometimes with square roots, there might be more conditions, but in this case, the expression inside the square root simplifies to a linear function. So once we have that, the only constraint is 10x -55 ≥ 0.

Alternatively, could there be cases where even though 10x -55 is non-negative, there's some other restriction? For example, if there was a denominator or something else, but here the function is just sqrt(...). So no, there's no other restrictions. So the domain is indeed x ≥ 11/2.

So the smallest real number x in the domain is 11/2. Therefore, the answer is 11/2, which is 5.5, and as a fraction, 11/2.

But let me check if maybe the original problem had some typos or something? The problem says "smallest real number x in the domain", so if the domain is [11/2, ∞), then the smallest number is 11/2.

Alternatively, if the problem had something else, but no. So I think that is correct. Therefore, I think the answer is 11/2.

**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 begin by analyzing the expression inside the square root. A square root is defined only when its argument is non-negative (i.e., $ \geq 0 $). Therefore, the domain of $ g(x) $ consists of all real numbers $ x $ for which

$$
(x - 3)^2 - (x - 8)^2 \geq 0.
$$

---

### Step 1: Simplify the expression inside the square root

We simplify the expression step by step using algebra or a difference of squares identity:

$$
(x - 3)^2 - (x - 8)^2 = (x - 3 - (x - 8))(x - 3 + (x - 8)).
$$

Compute each factor:

$$
x - 3 - x + 8 = 5,\quad x - 3 + x - 8 = 2x - 11.
$$

So the expression simplifies to:

$$
(x - 3)^2 - (x - 8)^2 = 5(2x - 11) = 10x - 55.
$$

---

### Step 2: Determine when the expression is non-negative

We now need to find the values of $ x $ such that:

$$
10x - 55 \geq 0.
$$

Solving the inequality:

$$
10x \geq 55 \Rightarrow x \geq \frac{55}{10} = \frac{11}{2}.
$$

---

### Step 3: Interpret the result

This tells us that the domain of $ g(x) $ is:

$$
\left[\frac{11}{2}, \infty\right).
$$

Hence, the smallest real number in this domain is:

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

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 $x\ge \frac{8+3}{2} = \boxed{\frac{11}{2}}$.