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.

  • $\begingroup$ even the littlest help would be appreciated, it is sort of urgent as i need to complete this question so i can answer the others and it will take me a while to answer them so i need help urgently $\endgroup$ – user215346 Feb 14 '15 at 4:21
  • $\begingroup$ This is not the answer I get. $\endgroup$ – MPW Feb 14 '15 at 5:33
  • $\begingroup$ Like @MPW, your given answer is not the one I get. Also, what do you mean by "the maximum width and length of the running facility"? Your conditions give the exact width and length: what changes for you to maximize? $\endgroup$ – Rory Daulton Feb 14 '15 at 13:13
  • $\begingroup$ A track facility is not a curve but a band having a certain width. The inner boundary is essentially shorter than the outer boundary. Where are the $400$ m to be measured? A figure would help greatly. $\endgroup$ – Christian Blatter Feb 14 '15 at 13:27

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

enter image description here

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?


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


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