Rectangular parallelepiped of greatest volume for a given surface area S

Question

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.

Answer 1

your procedure is correct if not for the slight differential errors in $\frac{dz}{dx} and \frac{dz}{dy}$ which has, in turn, resulted in all the wrong values. I followed your procedure, and these are the expressions and values I get for the differentials $$\begin{matrix} Z_x=\frac{-(k+y^2)}{(x+y)^2}&Z_y=\frac{-(k+x^2)}{(x+y)^2}\\ Z_{xx}=\frac{2(k+y^2)}{(x+y)^3}&Z_{yy}=\frac{2(k+x^2)}{(x+y)^3}\\ Z_{xy}=\frac{2(k-xy)}{(x+y)^3}\\ \end{matrix}$$ using these the double differential values of function V that you get will be $$V_{xx}=-a\;\;\;\;V_{yy}=-a\;\;\;V_{xy}=-a/2$$ I have ommitted writing the expressions for these as they were pretty long

Answer 2

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.