Building a new track – 3

Let $x$ be the length of the straight and $d$ be the diameter of the semicircles.

The inner edge of the track is $400 = 2x + \pi d$ and the area of the field is $A = xd$ .

Combining gives $A = -\frac{2}{\pi}x^2 + \frac{400}{\pi}x$ and completing the square gives $A = 10000 - \frac{2}{\pi}(x - 100)^2$ .

Therefore, the maximum area is $100000$ when $x - 100 = 0$ , or when $x = \boxed{100}$ meters.

If the length of the straights and the curves' diameters are $s$ and $c$ respectively, then we can set up the following equations

$\begin{aligned} 400 & = 2 \cdot s + 2 \cdot \frac 12 \cdot \pi c = 2s + \pi c\\ c & = \frac {400 - 2s}{\pi} \end{aligned}$

and then

$\begin{aligned} A_{field} & = s \cdot c \\ & = s \cdot \frac {400-2s}{\pi} \\ & = \frac {400}{\pi} s - \frac 2{\pi} s^2 \end{aligned}$

This is a parabola that is opened to the bottom, so its vertex will be the maximum we are looking for. To find it, let's differentiate with respect to $g$ and set the derivative equal to 0.

$\frac {dA}{dg} = \frac {400}{\pi} - \frac {4}{\pi} s$

$\frac {400}{\pi} - \frac {4}{\pi} s = 0 \Leftrightarrow \boxed{s=100}$

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This means that the maximum area will be reached when all four parts of the track are each 100m long, which is quite nice.