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?
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 $r=h$.
Let $r$ be the radius of the cylindrical can and $h$ be its height. Then we have $A = \pi r^2+2\pi r h$ fixed, as that is the total area. You will now need to determine the dimensions $r, h$ of the largest volume can, in terms of the known $A$.
One way to do this is to write $h$ in terms of $r$ and other known elements, as $h = \dfrac{A-\pi r^2}{2\pi r}$, and use this in the formula of $V$, i.e. $V = \pi r^2h = \frac12r(A-\pi r^2)$. Now we have $V$ as a function of one variable $r$, and calculus can be used to find its maximum.
$$\frac{dV}{dr} = \frac{d}{dr}\left(\frac12rA - \frac12 \pi r^3 \right)= \frac12(A-3\pi r^2)$$ For an extremum the above is zero, i.e. $r^2 = \dfrac{A}{3\pi}.\;$ As $\dfrac{dV^2}{dr^2}=-\frac32 \pi r < 0$, we are assured this is a maximum.
Now $h = \sqrt{\dfrac{A}{3\pi}}$, $V_{max} = \sqrt{\dfrac{A^3}{27\pi}}$ etc. can be worked out.
Let's look at the following Lagrangian:
$$\mathcal{L}(r,h,\lambda)=\pi r^2h-\lambda\left(\pi r^2+2\pi rh-A\right)$$
Let's solve for $\nabla \mathcal{L}(r,h,\lambda)=0$. We get three equations:
$$\pi r^2-2\pi \lambda r=0$$ $$2\pi(h-\lambda) r-2\pi\lambda h=0$$ $$\pi r^2+2\pi rh-A=0$$
Using the first equation, we get that $\lambda=\displaystyle\frac{r}{2}$ (since $r\neq 0$). Plugging this into the second equation, we get the following: $$r^2=rh$$
This implies that $r=h$ (since $r\neq0$). So the volume of the cylinder is maximized when the radius is equal to the height.
