Problem about length and width of a running facility

Question

Jacaranda Secondary College is planning to develop a $400$ metre running track facility in an unused area of the college. The rectangular site available is $100$ metres wide and $180$ metres long. The running track consists of two straight sections each $85$ metres long and two semicircles at each end. What will be the maximum width and length of the running facility.

The answer is $93.53\;\mathrm{m} \times 178.52 \;\mathrm{m}$.

I have tried figuring out the area of the running track, $2336.8$ metres, and the total area, $18000$ metres. I tried a few other things too but I didn't quite get the answer I need.

Answer 1

Use this diagram, which represents an idealization of your problem (the track has zero width, and so on):

Let us say that the "width" of the track is $d$, so the radii of the semicircles at the end are $\frac d2$. The straight sections have length $85$.

The condition is that the length of the track is $400$. Each semicircle has the length $\frac{\pi d}2$, so we get the equation

$$2\cdot 85 + 2\cdot\frac{\pi d}2=400$$ $$170+\pi d=400$$ $$\pi d=230$$ $$d=\frac{230}{\pi}\approx 73.21$$

Then the length $L$ and width $W$ are given by

$$W=d\approx 73.21$$ $$L=85+2\cdot\frac d2\approx 158.21$$

Is this what you needed?

Answer 2

Neglecting track width curved length = 400- 2*85 = 230 . Divide this by $ \pi$ we get 73.21 as maximum width of plan of track. And plan length is 85 + 73.21 = 158.21