<Question>
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?

[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));
[/asy]
</Question>

<think>
<Plan_1>
Okay, let me try to figure out this problem. So, there's a rectangular field enclosed in an athletic track. The track has two straight edges of the field and two semicircular ends. The total length of the track is 400 meters, and we need to find the largest possible area of the rectangular field.

Hmm, let's start by visualizing the track.
From the Asymptote code, it seems like the track is made up of two straight sides (which are the edges of the rectangle) and two semicircular ends. So, the entire track would form a sort of oval shape around the rectangle.

<prompt_1.1> Visualize the track and understand the problem setup as described in the above plan. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
The length of the track is 400 meters. I think the length here refers to the total distance around the track, which is like the perimeter of the entire athletic field. But since the track has two straight parts and two curved parts, the total length would be the sum of the lengths of the straight parts and the circumference of the curved parts.

Wait, but actually, in athletics tracks, the total length is the distance you run around the track, which includes going along the straight and then around the curve. So, maybe the 400 meters is the total length of the track, which would be the perimeter of the outer edge of the track. But in this case, the track is enclosing the rectangular field. So, the field is inside the track. So, the track's perimeter is 400 meters, which includes the two straight sides (which are the same as the length of the field) and the two semicircular ends (which together make a full circle).

Wait, let me confirm. If the track has two straight edges (the sides of the rectangle) and two semicircular ends, then the total perimeter of the track would be the sum of the lengths of the two straight sides plus the circumference of the two semicircles. But two semicircles make a full circle. So, the perimeter of the track is 2 times the length of the straight side plus 2 times pi times the radius of the semicircles.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But the problem is asking for the largest possible area of the field. The field is rectangular, so its area would be length times width. The length of the field would be the same as the straight sides of the track, right? Because the track has two straight edges that match the length of the field. Then the width of the field would be related to the diameter of the semicircular ends.

Wait, but the semicircular ends have a radius, let's call it r. So, the width of the field would be twice the radius, since the diameter is 2r. So, the area of the field would be length times width, which is L * 2r.

But we need to express this in terms of one variable to find the maximum. Let's see. The perimeter of the track is 400 meters.

<prompt_2.1> Set up the problem in terms of variables and constraints as described in the above plan. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
So, the perimeter of the track is the sum of the two straight sides (each of length L) and the two semicircular ends (each with circumference pi * r, so together 2 * pi * r). Therefore, the equation is:

2L + 2πr = 400.

But we need to find the maximum area, which is L * 2r. So, we can express one variable in terms of the other using the perimeter equation.

Let me write that down:

Perimeter: 2L + 2πr = 400.

Area: A = L * 2r.

We need to maximize A with respect to L and r, given the constraint from the perimeter.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
First, let's solve the perimeter equation for one variable. Let's solve for L:

<prompt_3.1> Solve the perimeter equation for L and substitute into the area formula as described in the above plan. </prompt_3.1>
-----

Wait, but is there any other constraint?

<prompt_3.2> Consider any other constraints and confirm the solution as described in the above plan. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
2L + 2πr = 400

Divide both sides by 2:

L + πr = 200

Therefore, L = 200 - πr.

Now, substitute this into the area formula:

A = L * 2r = (200 - πr) * 2r = 2r*(200 - πr) = 400r - 2πr².

So, the area is a quadratic function in terms of r: A(r) = -2πr² + 400r.

To find the maximum area, we can treat this as a quadratic equation. Since the coefficient of r² is negative, the parabola opens downward, so the vertex is the maximum point. The vertex of a quadratic ax² + bx + c is at x = -b/(2a).

Here, a = -2π, b = 400. Therefore, the value of r that maximizes the area is:

r = -b/(2a) = -400/(2*(-2π)) = -400 / (-4π) = 100/π.

So, r = 100/π meters.

Then, substitute back into the equation for L:

L = 200 - πr = 200 - π*(100/π) = 200 - 100 = 100 meters.

Therefore, the maximum area is A = L * 2r = 100 * 2*(100/π) = 100 * 200/π = 20000/π square meters.

Wait, but let me check this again. If L = 100 meters and r = 100/π, then the area is 100 * 2*(100/π) = 20000/π. That's approximately 20000 / 3.1416 ≈ 6366.2 square meters.
</execution_3.1>

<execution_3.2>
The problem states that the track consists of two edges of the field and two semicircles. So, the edges of the field are the straight parts of the track, so the length of the track's straight edges are equal to the length of the field. The semicircular ends have a diameter equal to the width of the field. So, yes, the width of the field is 2r, so the radius r is half the width. Therefore, the relationship between the field's width and the radius is direct. So, if the radius is larger, the width is larger, but the straight part (length) would have to be smaller because the total perimeter is fixed.

Therefore, the maximum area occurs when L = 2πr, which would mean that the field is a square in some sense, but since the track is a combination of straight and curved parts, the maximum area is achieved when the length equals the circumference of the circle formed by the two semicircles. That is, when L = 2πr. In that case, substituting back into the perimeter equation:

2L + 2πr = 400.

If L = 2πr, then 2*(2πr) + 2πr = 400 => 4πr + 2πr = 6πr = 400 => r = 400/(6π) ≈ 21.22 meters. Then L = 2πr ≈ 2π*(400/(6π)) = 400/3 ≈ 133.33 meters. Then the area would be L*2r = (400/3)*(800/(6π))*(2π) ? Wait, maybe I miscalculated. Wait, if L = 2πr, then area A = L * 2r = 2πr * 2r = 4πr². But if L = 2πr, then from the perimeter equation:

2L + 2πr = 400 => 2*(2πr) + 2πr = 6πr = 400 => r = 400/(6π) = 200/(3π).

Then A = 4πr² = 4π*(200/(3π))² = 4π*(40000)/(9π²) = (160000)/(9π). Which is approximately 160000/(28.2743) ≈ 5663.37 meters². But if we take L = 200 meters and r = 100/π ≈ 31.83 meters, then the area would be 200*(200/π) ≈ 200*63.66 ≈ 12,732 meters², which is larger. Wait, so my previous reasoning was wrong.

Wait, so when I set L = 2πr, that led to a smaller area. But when I took L = 200 and r = 100/π, which is a different ratio, the area is larger. Therefore, the maximum occurs when L is 200 meters and r is 100/π, leading to an area of 200*(200/π) = 40,000/π ≈ 12,732.395 meters². So that's the maximum.

But why does this happen? Because when we take the derivative and set it to zero, we find that the critical point is at L = 200, which gives a larger area than when we set L = 2πr. Therefore, the maximum area occurs when the length is 200 meters and the radius is 100/π meters. Therefore, the maximum area is 40,000/π square meters.

I think this is correct.
</execution_3.2>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But let me verify this with calculus, just to be sure. The area function is A(r) = -2πr² + 400r.

<prompt_4.1> Verify the solution using calculus as described in the above plan. </prompt_4.1>
-----

Based on execution_3.2:
But let me check once more. Let me suppose that the track is 400 meters.

<prompt_4.2> Perform a final verification and conclude the solution as described in the above plan. </prompt_4.2>
</Plan_4>

<Execution_4>
<execution_4.1>
To find the maximum, take the derivative of A with respect to r:

dA/dr = 400 - 4πr.

Set derivative equal to zero:

400 - 4πr = 0

4πr = 400

r = 400/(4π) = 100/π.

So, r = 100/π meters, which matches the previous result. Then L = 200 - πr = 200 - π*(100/π) = 200 - 100 = 100 meters. Then the area is L * 2r = 100 * 2*(100/π) = 20000/π ≈ 6366.197 meters².

Therefore, the maximum area is 20000/π square meters. But let me confirm if this is indeed the maximum. Since the coefficient of r² is negative, the function is concave down, so the critical point we found is indeed a maximum.

Alternatively, we can check the second derivative. The second derivative of A with respect to r is d²A/dr² = -4π, which is negative, confirming that the critical point is a maximum. Therefore, the maximum area is 20000/π.

But let me check if there are any other constraints. For example, the radius must be positive, and L must be positive. So, r > 0 and L = 200 - πr > 0. Therefore, 200 - πr > 0 => r < 200/π ≈ 63.66 meters. Since r = 100/π ≈ 31.83 meters, which is less than 63.66, so that's valid. So, the solution is feasible.

Therefore, the largest possible area of the field is 20000/π square meters. To write it as a boxed expression, it would be \boxed{\dfrac{20000}{\pi}}.

Wait, but let me check once more with different values. Suppose r = 20 meters.

<prompt_4.2>
**Final Answer**
The largest possible area of the field is \boxed{\dfrac{20000}{\pi}} square meters.
</execution_4.1>

<execution_4.2>
The total perimeter of the track is 400 meters. The track is made up of two straight sides (each of length L) and two semicircular ends (each of circumference πr, so together 2πr). Therefore, 2L + 2πr = 400. Then, the area of the field is L*2r. So, if we solve for L, we get L = (400 - 2πr)/2 = 200 - πr. Then, Area = (200 - πr)*2r = 400r - 2πr². Taking derivative: dA/dr = 400 - 4πr. Setting to zero: 400 - 4πr = 0 => r = 400/(4π) = 100/π. So, yes, that's correct. Therefore, r = 100/π, L = 200 - π*(100/π) = 200 - 100 = 100. Therefore, the area is 100*(200/π) = 20000/π. So, approximately 6366.197 meters².

But wait, maybe the problem is considering the track as the outer perimeter. Wait, but the field is inside the track. So, the length of the track's straight part is the same as the length of the field? Or is the length of the track's straight part longer?
</execution_4.2>
</Execution_4>

<Final_answer>
Therefore, the largest possible area of the field, in square meters, is $\boxed{\frac{20000}{\pi}}$.
</Final_answer>
</think>