Cost of a can (check my answer please) Question goes...A can is in the shape of a circular cylinder is required to have a volume of 750 cubic centimeters.  The top and bottom are made of material that costs 8 cents per square centimeter while the sides are made of a material that costs 5 cents per square centimeter.  Express the total cost $C$ of the material as a function of the radius $r$ of the cylinder.  For what value of $r$ is the cost $C$ the least?
Here's what I did but for some reason I'm not too confident with it.
Top and bottom: 2 circles; area of circle is $\pi r^2$, so total area is $2\pi r^2$.
Side: Lateral surface area is $2\pi rh$.
$C=2\pi r^2(.08)+2\pi rh(.05)$.
$V=\pi r^2h$
$750=\pi r^2h$
$h=\frac{750}{\pi r^2}$
So now $C=2\pi r^2(.08)+2\pi r\left(\frac{750}{\pi r^2}\right)$
$C=.16\pi r^2+.1\pi r\left(\frac{750}{\pi r^2}\right)$
$C=.16\pi r^2+\frac{75}{r}$ (multiply $16\pi r^2$ by $r$ to get same denominator)
$C=\frac{.16\pi r^3+75}{r}$.
I put this in my graph and got the minimum to be \$4.21 but that seems too high to be an answer.

In order to prove your answer you need to find first derivative of $C(r)$ and than calculate $r$ from equation $C'_r=0$  (in that point function $C(r)$ has minimum)
$C'_r=\frac{(0.16\pi r^3+75)'r-r'(0.16\pi r^3+75)}{r^2}\Rightarrow C'_r=\frac{0.32\pi r^3-75}{r^2}$ , so $C'_r=0 \Rightarrow 0.32\pi r^3-75=0 \Rightarrow r^3=\frac{75}{0.32\pi} \Rightarrow r=\sqrt[3]{\frac{75}{0.32\pi}}\Rightarrow r=4.21 cm$ , And if we substitute this value into $C(r)$ we get minimal cost of can which is $26.72$ cents.
