http://209.85.48.8/9963/162/upload/p2222876.jpg

Not sure how to do the last bit, help please.

Note: The cover is sometimes referred as a lid for some reason.

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- Dec 2nd 2010, 10:52 PMCthulFinding the minimum cost for a semi-circular cover
http://209.85.48.8/9963/162/upload/p2222876.jpg

Not sure how to do the last bit, help please.

Note: The cover is sometimes referred as a lid for some reason. - Dec 2nd 2010, 11:04 PMUnknown008
You didn't show the right equation for part (c)... =/

Area of two end sections = $\displaystyle \pi r^2$

Area of body of cover = $\displaystyle \dfrac12 2\pi r h = \pi rh$[/tex]

Cost of end sections = $\displaystyle 8\pi r^2$

Cost of body = $\displaystyle 4\pi rh$

Total cost = $\displaystyle 8\pi r^2 + 4 \pi rh$

Now, $\displaystyle h = \dfrac{16}{\pi r^2}$

Hence, cost = $\displaystyle 8\pi r^2 + 4 \pi r\left( \dfrac{16}{\pi r^2}\right) = 8\pi r^2 + \left( \dfrac{64}{ r}\right) $

4. I hope that you know some calculus... you'll have to find $\displaystyle \dfrac{dC}{dr} = 0$ then solve for r. - Dec 2nd 2010, 11:23 PMCthul
I know how to do calculus. Thanks.