The greatest area for a rectangle on a track field.

Question

An athletic field with a perimeter of 0.25 miles consists of a rectangle with a semicircle at each end, as shown below. Find the dimensions that yield the greatest possible area for the rectangular region.

This is the work that I did below. I was wondering if this was the greatest possible area for the rectangle below.

score 0 · Accepted Answer · answered Dec 11 '18 at 01:47

0

Near the end of page $1$, you wrote $r=\frac{2}{16\pi}$ when you meant to say $2r=\frac{2}{16\pi}$
Once we found out that $r=\frac{1}{16\pi}$, we can compute $$l=\frac18 - \pi r= \frac18 -\frac1{16}=\frac1{16}$$ directly without finding $A$ explicitly.

answered Dec 11 '18 at 01:47

Siong Thye Goh

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why would it be 2r? – mjj Dec 11 '18 at 01:49
you wrote $r=\frac2{16\pi}$ and then you wrote $w=\frac{1}{8\pi}$? – Siong Thye Goh Dec 11 '18 at 01:53
I got r to equal 1/16pi. I then multiplied this by 2 because of 2r=w. I then got 1/8pi for w. – mjj Dec 11 '18 at 01:55
Great, do not write $r = \frac1{16\pi} \times 2$, you can write $r \times 2 = \frac1{16\pi} \times 2$ or $2r = \frac1{16\pi} \times 2$. – Siong Thye Goh Dec 11 '18 at 01:56
so the 'l' and the 'w' are still correct? – mjj Dec 11 '18 at 01:57
yup, just a minor (but very crucial in the long run) writing issue and longer working, it's fine. – Siong Thye Goh Dec 11 '18 at 01:58

The greatest area for a rectangle on a track field.

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