Finding the minimum cost for a semi-circular cover

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December 2nd 2010, 10:52 PM
Cthul

Finding 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.
December 2nd 2010, 11:04 PM
Unknown008

You didn't show the right equation for part (c)... =/

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

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

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

Cost of body = $4\pi rh$

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

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

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

4. I hope that you know some calculus... you'll have to find $\dfrac{dC}{dr} = 0$ then solve for r.
December 2nd 2010, 11:23 PM
Cthul

I know how to do calculus. Thanks.

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