A (right circular) cylindrical can has a volume of 60π cubic inches. Suppose that the metal used for the top and bottom of the can costs 4 cents per square inch, while material for the side of the can costs only 2 cents per square inch.
What is the DIAMETER of the can that has minimum cost?
The volume of a cylinder is
$$ V=\pi r^2h. $$
The surface area of a cylinder is
$$ A=2\pi r^2+2\pi rh. $$
In this problem, we seek to minimize $f$ subject to $g$, where
$$\begin{align} f(r,h)&=8\pi r^2+4\pi rh,\quad\text{and}\\ g(r,h)&=\pi r^2h=60\pi. \end{align}$$
Using the method of Lagrange multipliers, we have that:
$$\begin{align} \nabla f&=\left\langle16\pi r+4\pi h,4\pi r\right\rangle,\quad\text{and}\\ \nabla g&=\left\langle2\pi rh,\pi r^2\right\rangle; \end{align}$$
$$\begin{align} 16\pi r+4\pi h&=\lambda2\pi rh,\\ 4\pi r&=\lambda\pi r^2,\\ \pi r^2h&=60\pi; \end{align}$$
$$\begin{align} h&=16/\lambda,\\ r&=4/\lambda,\\ \lambda&=4/\sqrt[3]{15}. \end{align}$$
Therefore, $r=\sqrt[3]{15}$, and $d=2\sqrt[3]{15}$, as required.