Not sure how to do the last bit, help please.
Note: The cover is sometimes referred as a lid for some reason.
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.