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.


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.
