Answer
(a) the box $B$ is not a cube as in the solution to Exercise 50. We expect the base dimensions ($x$ and $y$) would be the same, but the height $z$ would be shorter.
(b) the dimensions of $B$ is $\left( {x,y,z} \right) = \left( {{{\left( {2V} \right)}^{1/3}},{{\left( {2V} \right)}^{1/3}},{{\left( {\frac{V}{4}} \right)}^{1/3}}} \right)$.
The smallest possible surface area is $S\left( {{{\left( {2V} \right)}^{1/3}},{{\left( {2V} \right)}^{1/3}}} \right) = 3{\left( {2V} \right)^{2/3}}$.
Work Step by Step
(a) Since the box has no top, $B$ is not a cube as in the solution to Exercise 50. The volume is fixed but the surface area is less than the one in Exercise 50. We expect the base dimensions ($x$ and $y$) would be the same, but the height $z$ would be shorter.
(b) Let the volume of $B$ be fixed: $V=x y z$, where $V$ is a constant.
Since $B$ has no top, the surface area of the rectangular box is $S = xy + 2xz + 2yz$.
Using $V=x y z$, we can express $S$ as a function of $x$ and $y$:
$S\left( {x,y} \right) = xy + 2x\left( {\frac{V}{{xy}}} \right) + 2y\left( {\frac{V}{{xy}}} \right) = xy + \frac{{2V}}{y} + \frac{{2V}}{x}$
The partial derivatives are
${S_x} = y - \frac{{2V}}{{{x^2}}}$, ${\ \ \ }$ ${S_y} = x - \frac{{2V}}{{{y^2}}}$
${S_{xx}} = \frac{{4V}}{{{x^3}}}$, ${\ \ \ }$ ${S_{yy}} = \frac{{4V}}{{{y^3}}}$, ${\ \ \ }$ ${S_{xy}} = 1$
We find the critical points of $S$ by solving ${S_x} = 0$ and ${S_y} = 0$:
${S_x} = y - \frac{{2V}}{{{x^2}}} = 0$, ${\ \ \ }$ ${S_y} = x - \frac{{2V}}{{{y^2}}} = 0$
From the first equation we obtain $y = \frac{{2V}}{{{x^2}}}$. Substituting it in the second equation gives
$x - \frac{{{x^4}}}{{2V}} = 0$
$x\left( {1 - \frac{{{x^3}}}{{2V}}} \right) = 0$
So, the solutions are $x=0$ and $x = {\left( {2V} \right)^{1/3}}$. However, $y$ is undefined if $x=0$. Thus, there is only one critical point $\left( {{{\left( {2V} \right)}^{1/3}},{{\left( {2V} \right)}^{1/3}}} \right)$.
Using $V=x y z$, we obtain the size of the box $\left( {x,y,z} \right) = \left( {{{\left( {2V} \right)}^{1/3}},{{\left( {2V} \right)}^{1/3}},{{\left( {\frac{V}{4}} \right)}^{1/3}}} \right)$.
We evaluate ${S_{xx}}$ and the discriminant $D$ at $\left( {x,y} \right) = \left( {{{\left( {2V} \right)}^{1/3}},{{\left( {2V} \right)}^{1/3}}} \right)$:
${S_{xx}} = \frac{{4V}}{{{{\left( {{{\left( {2V} \right)}^{1/3}}} \right)}^3}}} = 2 > 0$
$D = {S_{xx}}{S_{yy}} - {S_{xy}}^2 = 3 > 0$
Since ${S_{xx}} > 0$ and $D>0$, by Theorem 2, $S\left( {{{\left( {2V} \right)}^{1/3}},{{\left( {2V} \right)}^{1/3}}} \right) = 3{\left( {2V} \right)^{2/3}}$ is the local minimum of $S$.
Hence, the dimensions of $B$ is $\left( {x,y,z} \right) = \left( {{{\left( {2V} \right)}^{1/3}},{{\left( {2V} \right)}^{1/3}},{{\left( {\frac{V}{4}} \right)}^{1/3}}} \right)$. Notice that $z$ is shorter as predicted in part (a). Thus, $B$ is not a cube.
