## Problem on finding the rectangular prism of maximal volume Of the rectangular prisms with surface area A, which has maximal volume?
• ## Solution We observe that this is a constrained optimization problem: we are seeking to maximize the volume of a rectangular prism with a constraint on its surface area. First we sketch the prism and introduce variables for its dimensions $x,y,z$.   The volume and surface area of the prism are \begin{align}
\text{Volume} &= x y z\\
\text{Area} &= 2 xy + 2 xz + 2 yz
\end{align} Thus, we are trying to maximize $xyz$ subject to $2xy + 2xz + 2yz = A$. Recall that We identify $f(x,y,z) = xyz$ and $g(x,y,z) = 2xy + 2xz + 2yz - A$. We form the augmented function $$L(x,y,z, \lambda) = xyz - \lambda( 2 x y + 2 x z + 2 yz - A).$$ Solving each of the first three equations for $\lambda$: $$\lambda = \frac{yz}{2(y+z)} = \frac{xz}{2(x+z)} = \frac{xy}{2(x+y)}.$$ These equations are difficult to solve for $x,y,z$ because they are nonlinear. Each term becomes simpler if expressed in terms of the reciprocals of $x,y,z$. Solving each of the first three equations for $\lambda$: $$\lambda = \frac{1}{2(\frac{1}{y}+\frac{1}{z})} = \frac{1}{2(\frac{1}{x}+\frac{1}{z})} = \frac{1}{2(\frac{1}{x}+\frac{1}{y})}.$$ Thus, we get $\frac{1}{y}+\frac{1}{z} = \frac{1}{x}+\frac{1}{z} = \frac{1}{x}+\frac{1}{y}$. We simplify to $\frac{1}{x} = \frac{1}{y} = \frac{1}{z}$. Hence $x=y=z$ and the rectangular prism of maximal volume is a cube.