Rectangular parallelepiped of greatest volume for a given surface area S 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.

 A: 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.
A: 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 
