Consider a right circular cylinder of radius $r$ and height $h$. It has volume $V=\pi r^2 h$ and area $A=2\pi r (r+h)$. We are to use Lagrange multipliers to prove the maximum volume with given area is $$ V= \frac{1}{3}\sqrt{\frac{A^3}{6\pi}}$$
Here is my attempt. We set up: $$ \mathcal{L} = \pi r^2h – \lambda(2 \pi r^2+2\pi rh)$$ Follows: $$ \mathcal{L}_h = \pi r^2 – \lambda(2\pi r) = 0 $$ $$ \mathcal{L}_r = 2\pi rh – \lambda(4 \pi r+2\pi h) = 0 $$ Set these two equal to each other and noting that $$\mathcal{L}_\lambda = 2\pi r^2 + 2\pi r h = 0 $$ We get $$\boxed{2h=r}$$
Forgive me if I have done something obviously incorrect, but we square the resulting $V$ and cube the resulting $A$ in order to eliminate $r$. Setting $x$ as an arbitrary coefficient. $$V^2 = xA^3$$ $$ \frac{\pi^2 r^6}{4} = x27\pi^3r^6 \implies x = \frac{1}{108\pi} $$ and so $$V= \sqrt{\frac{A^3}{108\pi}} = \frac{1}{6}\sqrt{\frac{A^3}{3\pi}}$$ Painfully close, but not quite the answer required. Help is much appreciated
Best Answer
Check the step before $2h=r$, i get $h=2r$ with the same $\mathcal{L}_h$ and $\mathcal{L}_r$