Hello Scott:
The objective function (that which we wish to optimize) is the area of the rectangle, and is as follows:
$$f(r,x)=2rx$$
And if we let $P$ be the perimeter of the field, we may state the constraint:
$$g(r,x)=2x+2\pi r-P=0$$
Now, we have two ways to proceed. The first is to solve the constraint for one of the two variables $r$ or $x$, and then substitute into the objective function so that we have a function in one variable and then optimize by differentiation. Sol if we solve the constraint for $r$, we obtain:
$$r=\frac{P-2x}{2\pi}$$
And the substituting into the objective function, we obtain:
$$f(x)=2\left(\frac{P-2x}{2\pi}\right)x=\frac{x(P-2x)}{\pi}$$
Now, at this point we see we have a quadratic function opening downwards and we could use a precalculus technique to find the maximum value. We see the roots are:
$$x=0,\,\frac{P}{2}$$
We know then that the axis of symmetry lies midway between these roots, and so the axis of symmetry is at :
$$x=\frac{P}{4}$$
And so we have:
$$r=\frac{P-2\left(\frac{P}{4}\right)}{2\pi}=\frac{P}{4\pi}$$
Now, let's try differentiation. Recall:
$$f(x)=\frac{x(P-2x)}{\pi}=\frac{1}{\pi}\left(Px-2x^2\right)$$
Hence, differentiating with respect to $x$, and equating the result to zero to obtain the critical value(s), we obtain:
$$f'(x)=\frac{1}{\pi}\left(P-4x\right)=0\implies x=\frac{P}{4}$$
And like before, we find:
$$r=\frac{P}{4\pi}$$
We see also that:
$$f''(x)=-\frac{4}{\pi}<0$$
Since the objective function is concave down everywhere, we know the critical value is at the global maximum.
Now, let's examine a multivariable method: Lagrange multipliers. Recall, we have:
The objective function:
$$f(r,x)=2rx$$
Subject to the constraint:
$$g(r,x)=2x+2\pi r-P=0$$
Hence, we obtain the system:
$$2x=\lambda(2\pi)$$
$$2r=\lambda(2)$$
Hence, this implies:
$$\lambda=\frac{x}{\pi}=r$$
Substituting for $r$ into the constraint, we obtain:
$$2x+2\pi\left(\frac{x}{\pi}\right)-P=0$$
$$4x=P$$
$$x=\frac{P}{4}\implies r=\frac{P}{4\pi}$$
So, given that the objective function is zero for $x=0$, we may conclude that this critical point is at a maximum.
Thus, we have shown in various ways that:
$$f_{\max}=f\left(\frac{P}{4},\frac{P}{4\pi}\right)$$
Using the given value $P=400\text{ m}$, we then may conclude that the rectangular portion of the field is maximized for:
$$x=\frac{400\text{ m}}{4}=100\text{ m}$$
$$r=\frac{400\text{ m}}{4\pi}=\frac{100}{\pi}\text{ m}$$
This means that the straight portions of the track are equal in length to the curved portions.
