I am trying to find the rectangular parallelepiped of greatest volume for a given surface area S using Lagrange's method.
I tried solving by myself but at x=y=z = a, I am not getting maximum volume but minimum volume.
I have attached the procedure done by myself in attached picture. Please help me here.
Best Answer
Don't try to perform a second derivative test in connection with Lagrange's method. The point you have found is clearly the maximum. Using the AGM inequality you have $$\root 3 \of{V^2}=\root 3\of{ab\cdot bc\cdot ca}\leq{ab+bc+ca\over3}={1\over6}S\ ,$$ with equality sign iff $ab=bc=ca$, i.e., iff $a=b=c$. It follows that $$V\leq\left({S\over6}\right)^{3/2}$$ with equality iff the parallelopiped is a cube with the given surface area.