Answer
We show that the rectangular box is a cube of dimensions $\left( {x,y,z} \right) = \left( {3,3,3} \right)$.
The smallest possible surface area is $S\left( {3,3} \right) = 54$.
Work Step by Step
From Figure 25 we have the volume $V = xyz$. It is required that the volume be fixed $V=27$ ${m^3}$. So, $V=x y z=27$.
From Figure 25 we obtain the surface area of the rectangular box $S = 2xy + 2xz + 2yz$.
Using $V=x y z=27$, we can express $S$ as a function of $x$ and $y$:
$S\left( {x,y} \right) = 2xy + 2x\left( {\frac{{27}}{{xy}}} \right) + 2y\left( {\frac{{27}}{{xy}}} \right) = 2xy + \frac{{54}}{y} + \frac{{54}}{x}$
The partial derivatives are
${S_x} = 2y - \frac{{54}}{{{x^2}}}$, ${\ \ \ }$ ${S_y} = 2x - \frac{{54}}{{{y^2}}}$
${S_{xx}} = \frac{{108}}{{{x^3}}}$, ${\ \ \ }$ ${S_{yy}} = \frac{{108}}{{{y^3}}}$, ${\ \ \ }$ ${S_{xy}} = 2$
We find the critical points of $S$ by solving ${S_x} = 0$ and ${S_y} = 0$:
${S_x} = 2y - \frac{{54}}{{{x^2}}} = 0$, ${\ \ }$ ${S_y} = 2x - \frac{{54}}{{{y^2}}} = 0$
From the first equation we obtain $y = \frac{{27}}{{{x^2}}}$. Substituting it in the second equation gives
$x - \frac{{{x^4}}}{{27}} = 0$
$x\left( {1 - \frac{{{x^3}}}{{27}}} \right) = 0$
So, the solutions are $x=0$ and $x=3$. However, $y$ is undefined if $x=0$. Thus, there is only one critical point $\left( {3,3} \right)$.
Using $V=x y z=27$ we obtain the size of the box $\left( {x,y,z} \right) = \left( {3,3,3} \right)$, which is a cube.
We evaluate ${S_{xx}}$ and the discriminant $D$ at $\left( {x,y} \right) = \left( {3,3} \right)$:
${S_{xx}} = 4 > 0$
$D = {S_{xx}}{S_{yy}} - {S_{xy}}^2 = 12 > 0$
Since ${S_{xx}} > 0$ and $D > 0$, by Theorem 2, $S\left( {3,3} \right) = 54$ is the local minimum of $S$.
Hence, the smallest possible surface area is a cube of dimensions $\left( {x,y,z} \right) = \left( {3,3,3} \right)$.
