See problem 7 and figure 9 in picture

enter image description here

What I've done so far:

Not sure if $P=2l+2w$ or just $l+2w$ (dashed line makes me think the latter)

$600=\pi r+l+2w$

$600=\pi r+2r+2w$

$w=\frac{600-\pi r-2r}{2}=300-\frac{\pi}{2}r-r$

$A=\frac{1}{2}\pi r^2+lw$

$A=\frac{1}{2}\pi r^2+2r(300-\frac{\pi}{2}r-r)=\frac{1}{2}\pi r^2+600r-\pi r^2-2r^2$

$\frac{dA}{dr}=\pi r +600-2\pi r-4r=600-\pi r-4r$

$r=\frac{600}{\pi+4}, l\approx 84.01$ (half the value it should be)

I'm not sure where I went wrong or if I went about it the wrong way. Thanks in advance

  • 2
    $\begingroup$ It looks right except at the end. What you call $l$ is $2r$. $\endgroup$ Dec 6, 2014 at 0:45

1 Answer 1


In your steps, you imply $l=2r$. Therefore, if $r = \frac{600}{\pi+4}$, then $l = 2r = \frac{1200}{\pi+4} \approx 168.029746$


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