## Problem on Lagrange multipliers Consider an open box with no top, as shown. The box has volume $32$ and dimensions $x,y,z$. Using Lagrange multipliers, find the dimensions of the box with minimal surface area.  • ## Solution We observe this is a constrained optimization problem: we are to minimize surface area under the constraint that the volume is 32. We now express the problem in terms of the variables $x,y,z$. Because the box is open, it has surface area $2xy + 2xz + yz$. The constraint on volume is that $xyz = 32$. Recall that To apply the above procedure, we have to write the problem in the form above. That is, we need to identify $f$ and $g$. We identify that $f(x,y,z) = 2xy + 2xz + yz$ and $g(x,y,z) = xyz-32$. Forming the augmented function, we get $$L(x,y,z, \lambda) = 2 xy + 2xz + yz - \lambda(xyz - 32).$$ Because these equations are nonlinear, we can not use linear algebra techniques to solve them. Instead, we try to make progress by writing equations for $\lambda$:
\begin{align}
\lambda &= \frac{2y+2z}{yz} \\
\lambda &= \frac{2 x+z}{xz} \\
\lambda &= \frac{2 x+y}{xy} \\
\end{align} These equations can be simplified so that each variable only appears once: \begin{align}
\lambda &= \frac{2y+2z}{yz} = \frac{2}{z} + \frac{2}{y} \\
\lambda &= \frac{2 x+z}{xz} = \frac{2}{z} + \frac{1}{x} \\
\lambda &= \frac{2 x+y}{xy} = \frac{2}{y} + \frac{1}{x}
\end{align} Equating the first two equations gives $\frac{2}{z} + \frac{2}{y} = \frac{2}{z} + \frac{1}{x}$, which implies $\frac{2}{y} = \frac{1}{x}$. Thus $y = 2 x$. Equating the last two equations gives $\frac{2}{z} + \frac{1}{x} = \frac{2}{y} + \frac{1}{x}$, which implies $\frac{2}{z} = \frac{2}{y}$. Thus $y=z$. We can substitute $y$ and $z$ into the constraint $xyz = 32$ in order to compute $x = 2$. Thus $y = z = 4$. We see that the minimum surface area is given by $x=2, y=4, z=4$.