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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.

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  • 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.

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  • $\begingroup$ why would it be 2r? $\endgroup$ – mjj Dec 11 '18 at 1:49
  • $\begingroup$ you wrote $r=\frac2{16\pi}$ and then you wrote $w=\frac{1}{8\pi}$? $\endgroup$ – Siong Thye Goh Dec 11 '18 at 1:53
  • $\begingroup$ I got r to equal 1/16pi. I then multiplied this by 2 because of 2r=w. I then got 1/8pi for w. $\endgroup$ – mjj Dec 11 '18 at 1:55
  • $\begingroup$ Great, do not write $r = \frac1{16\pi} \times 2$, you can write $r \times 2 = \frac1{16\pi} \times 2$ or $2r = \frac1{16\pi} \times 2$. $\endgroup$ – Siong Thye Goh Dec 11 '18 at 1:56
  • $\begingroup$ so the 'l' and the 'w' are still correct? $\endgroup$ – mjj Dec 11 '18 at 1:57

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