A manufacturer wants to make an open rectangular box of volume \(V=500\) cubic centimeters (\(\rm{cm}^3\)) using the least possible amount of material. Find the dimensions of the box.

Solution  Let \(x\) and \(y\) be the dimensions of the base of the box, and let \(z\) be the height of the box. Then \[ V=500=xyz\,\rm{cm}^{3}\qquad x>0\quad y>0\quad z>0 \]

The manufacturer wants to minimize the amount of material used, which equals the surface area \(S\) of the box. Since the box is open, \[ S=xy+2xz+2yz \]

If we solve the equation \(500=xyz\) for \(z\) and substitute \(z=\dfrac{500}{xy}\) into the formula for the surface area \(S\) of the box, we can express \(S\) as a function of two variables. \[ S=S( x,y) =xy+2x\!\left( \dfrac{500}{xy}\right) +2y\left( \dfrac{500 }{xy}\right) =xy+\frac{1000}{y}+\frac{1000}{x} \]

This is the function to be minimized. The partial derivatives of \(S\) are \[ S_{x}=y-\dfrac{1000}{x^{2}}\qquad \hbox{and} \qquad S_{y}=x-\dfrac{1000}{y^{2}} \]

Since \(x>0\) and \(y>0,\) the critical points satisfy the system of equations \[ \left\{\begin{array}{l} y-\dfrac{1000}{x^{2}}=0\\[3pt] x-\dfrac{1000}{y^{2}}=0 \end{array}\right. \]

Using substitution, we find \[ x=y=1000^{1/3}=10 \]

The second-order partial derivatives of \(S\) are \[ \begin{eqnarray*} S_{xx}&=&\dfrac{\partial }{\partial x}\left( y-\dfrac{1000}{x^{2}}\right) =\dfrac{2000}{x^{3}}\qquad S_{xy}= \dfrac{\partial }{\partial x}\left(x-\dfrac{1000}{y^{2}}\right) =1\\[4pt] S_{yy}&=&\dfrac{\partial }{\partial y}\left( x-\dfrac{1000}{y^{2}}\right) =\dfrac{2000}{y^{3}} \end{eqnarray*} \]

At the critical point \(( 10,10) \), \[ \begin{eqnarray*}\hspace{-1pc} &\hspace{-1.8pc}A=S_{xx}(10,10) =\dfrac{2000}{10^{3}}=2\qquad B=S_{xy}(10,10) =1\qquad C=S_{yy}(10,10) =\dfrac{2000}{10^{3}}=2 &\hspace{1.8pc}\\[4pt] &AC-B^{2}=3>0& \end{eqnarray*} \]

From the Second Partial Derivative Test, \(S\) has a local minimum at \((10,10)\).

But, we want to know where \(S\) attains its absolute minimum. Since the physical properties of the problem require that the absolute minimum exists, the local minimum is also the absolute minimum. Consequently, the dimensions (in centimeters) of the open box of volume \(V=500\,\rm{cm}^{3}\) that uses the least amount of material are \[ x=10\rm{cm}\qquad y=10\rm{cm}\qquad z=\dfrac{500}{100}=5\rm{cm} \]