Calculus of the cost of a can I have found an equation for the cost of a cylindrical can (with height $h$, radius $r$, cents $k$) to be $C(h,r)=k(2\pi r)(r+h)$ and I am trying to figure out for a fixed volume $V_0$, show that the minimum cost for the can occurs when $\frac {h}{r}=2$. Any help would be greatly appreciated.
 A: The fixed volume $V_0$ is
$$
V_0 = \pi r^2 h,
$$
which gives
$$
h\left(r\right) = \frac{V_0}{\pi r^2}.
$$
So take the derivative of
$$
C\left(r\right) = C\left(h\left(r\right),r\right) = 2 \pi k r \left(h\left(r\right)+r\right) =2 \pi k r \left(\frac{V_0}{\pi r^2}+r\right) = 2 k \left(\frac{V_0}{r}+\pi r^2\right)
$$
with respect to $r$ and set it to zero.
A: Here is how to do it:


*

*Express $V_0$ as a function of $r$ and $h$.

*Re-arrange this expression to get $h$ as a function of $r$ (and the constant $V_0$).

*Substitute this value for $h$ in your expression for $C(h,r)$.

*Differentiate the resulting cost function with respect to $r$.

*Set this derivative of the cost function to $0$, and solve for $r$.

*Check that this really gives a minimum, by calculating the second derivative at $r$ and showing that it is positive. (This step can be skipped if all you want is the answer. But in a formal proof, you do need to show somehow that you have a minimum.)


Let us know if you have trouble with any of these steps.
