A container in the shape of a rectangular box is open on top and has a fixed volume of 12 m$^3$. The material used to make the bottom of the container costs $3 per square meter, while the material used for the sides costs $1 per square meter. What dimensions will minimize the cost of material?

Solution Let $x$ be the width of the container, $y$ its depth, and $z$ its height. Then its volume is $xyz=12.$ The cost function $C=C(x,y,z)$, which is to be minimized, is $$C(x,y,z) =3xy+2xz+2yz$$

We use Lagrange multipliers.

Step 1 The objective is to minimize $C$ subject to the constraint $g(x,y,z)=xyz-12=0$.

Step 2 The functions $C$ and $g$ each have continuous partial derivatives.

Step 3 We solve the system of equations ${\nabla }C(x,y,z) =\lambda {\nabla }g(x,y,z)$ and $g(x,y,z) =0$. $$\left\{\begin{array}{lllll} 3y+2z &=&\lambda yz \qquad {\color{#0066A7}{\hbox{\(C_{x}(x,y,z) =\lambda g_{x}( x,y,z)\)}}} \enspace \enspace {\color{#0066A7}{\hbox{(\(1\))}}}\\ 3x+2z &=&\lambda xz \qquad {\color{#0066A7}{\hbox{\(C_{y}(x,y,z) =\lambda g_{y}( x,y,z)\)}}} \enspace \enspace {\color{#0066A7}{\hbox{(\(2\))}}}\\ 2x+2y &=&\lambda xy \qquad {\color{#0066A7}{\hbox{\(C_{z}(x,y,z) =\lambda g_{z}( x,y,z)\)}}} \enspace \enspace {\color{#0066A7}{\hbox{(\(3\))}}}\\ xyz-12 &=&0 \qquad \enspace \enspace \enspace {\color{#0066A7}{\hbox{\(g( x,y,z) =0\)}}}\quad \enspace \qquad \enspace \enspace \enspace{\color{#0066A7}{\hbox{(\(4\))}}} \end{array}\right.$$

Now we use the facts that $x>0$, $y>0$, and $z>0$ to solve the first three equations for $\lambda$, obtaining $$\lambda =\dfrac{3}{z}+\frac{2}{y}\enspace {\color{#0066A7}{\hbox{(\(5\))}}} \qquad \lambda =\frac{3}{z}+\frac{2}{x}\enspace {\color{#0066A7}{\hbox{(\(6\))}}} \qquad \lambda =\dfrac{2}{y}+\frac{2}{x}\enspace {\color{#0066A7}{\hbox{(\(7\))}}}$$

From these, we find that $$\begin{eqnarray*} y &=&x \qquad \enspace\enspace\enspace {\color{#0066A7}{\hbox{From equations (5) and (6)}}}\\ z &=&\dfrac{3}{2}x \qquad \enspace \enspace {\color{#0066A7}{\hbox{From equations (5) and (7)}}} \end{eqnarray*}$$

If we substitute these two equations into $xyz-12=0$, we find \[ \begin{eqnarray*} xyz-12& =&0 \\[4pt] x( x) \left( \dfrac{3}{2}x\right) -12& =&0 \\[4pt] \frac{3}{2}x^{3}& =&12 \\[4pt] x& =&2 \end{eqnarray*}

Then $y=2,$ $z=3,$ and the test point is $(2,2,3)$.

Step 4 The dimensions of the container that minimize the cost are 2 m by 2 m by 3 m. The minimum cost is $C(2,2,3)=3 ( 2) ( 2) +2 ( 2) (3) +2 ( 2) ( 3) =$36.$