Answer
The dimensions that will minimize the cost of the materials for the box is $\left( {x,y,z} \right) = \left( {1,1,8} \right)$.
Work Step by Step
We have the following costs:
$\begin{array}{*{20}{c}}
{}&{{\rm{Cost}}\left( {\rm{\$ }} \right)}\\
{{\rm{gold{-}plated{\ }top}}}&{120xy}\\
{{\rm{silver{-}plated{\ }bottom}}}&{40xy}\\
{{\rm{copper{-}plated{\ }sides}}}&{10\left( {2xz + 2yz} \right)}\\
{{\rm{Total{\ }cost}}\left( f \right)}&{160xy + 20xz + 20yz}
\end{array}$
The box has a fixed volume of $V=x y z=8$ ${m^3}$.
Using $x y z=8$, we can write the total cost $f$ as a function of $x$ and $y$:
$f\left( {x,y} \right) = 160xy + 20x\left( {\frac{8}{{xy}}} \right) + 20y\left( {\frac{8}{{xy}}} \right)$
$f\left( {x,y} \right) = 160xy + \frac{{160}}{y} + \frac{{160}}{x}$
The partial derivatives are
${f_x} = 160y - \frac{{160}}{{{x^2}}}$, ${\ \ \ }$ ${S_y} = 160x - \frac{{160}}{{{y^2}}}$
${f_{xx}} = \frac{{320}}{{{x^3}}}$, ${\ \ }$ ${f_{yy}} = \frac{{320}}{{{y^3}}}$, ${\ \ }$ ${f_{xy}} = 160$
We find the critical points of $f$ by solving ${f_x} = 0$ and ${f_y} = 0$:
${f_x} = 160y - \frac{{160}}{{{x^2}}} = 0$, ${\ \ }$ ${S_y} = 160x - \frac{{160}}{{{y^2}}} = 0$
From the first equation we obtain $y = \frac{1}{{{x^2}}}$. Substituting it in the second equation gives
$160x - 160{x^4} = 0$
$x\left( {1 - {x^3}} \right) = 0$
So, the solutions are $x=0$ and $x=1$. However, $y$ is undefined if $x=0$. Thus, there is only one critical point $\left( {1,1} \right)$.
Using $V=x y z=8$ we obtain the dimensions of the box $\left( {x,y,z} \right) = \left( {1,1,8} \right)$.
We evaluate ${f_{xx}}$ and the discriminant $D$ at $\left( {x,y} \right) = \left( {1,1} \right)$:
${f_{xx}} = 320$
$D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = 76800 > 0$
Since ${f_{xx}} > 0$ and $f>0$, by Theorem 2, $f\left( {1,1} \right) = 480$ is the local minimum of $f$. So, the minimum cost is $\$ 480$.
Hence, the dimensions that will minimize the cost of the materials for the box is $\left( {x,y,z} \right) = \left( {1,1,8} \right)$.
