A cylindrical can without a top is made using $A \text{ cm}^2$ of material. Find the dimensions that will maximize the volume of the can.
What I have done was similar to the question: Optimization with cylinder
And I came up with the same result of $$R=\sqrt[3]{\frac{V}{\pi}}$$
However, is it the same to find $$V\left(\sqrt[3]{\frac{V}{\pi}}\right)$$ to find the maximum?
Best Answer
We can find the maximum volume without using Calculus: \begin{align*} V^2 &= \frac{1}{4}r^2(A-\pi r^2)(A-\pi r^2)\\ &= \frac{1}{8\pi}2\pi r^2(A-\pi r^2)(A-\pi r^2) \\ &\leq \frac{1}{8\pi} \left(\frac{2\pi r^2 + A -\pi r^2 + A - \pi r^2}{3}\right)^3 \\ &= \frac{1}{8\pi}(8A^3/27) = \frac{A^3}{27\pi} \end{align*} by A.M - G.M inequality. Equality holds when $A - \pi r^2 = 2 \pi r^2$ or $3\pi r^2 = \pi r^2 + 2\pi rh$ or $r=h$.
Thus the maximum volume is $\sqrt{\frac{A^3}{27\pi}}$ and occurs when $r=h$.