A manufacturer needs to produce a cylindrical container with a capacity of $1000\ {\rm{cm}}^{3}$. The top and bottom of the container are made of material that costs $0.05 per square centimeter, while the sides of the container are made of material costing $0.03 per square centimeter. Find the dimensions that will minimize the company’s cost of producing the container.

Solution Figure 58 shows a cylindrical container and the area of its top, bottom, and lateral surfaces. As shown in the figure, we let $h$ denote the height of the container and $R$ denote the radius. The total area of the bottom and top is $2\pi\! R^{2}\ {\rm{cm}}^{2}$. The area of the lateral surface of the can is $2\pi\! Rh\ {\rm{cm}}^{2}$.

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The variables $h$ and $R$ are related. Since the volume of the cylinder is $1000\ {\rm{cm}}^{3}$, $$\begin{eqnarray*} V &=&\pi\! R^{2}h=1000 \\[3pt] h &=&\frac{1000}{\pi\! R^{2}} \end{eqnarray*}$$

The cost $C$, in dollars, of manufacturing the container is $$C=(0.05)({2\pi\! R^{2}})+(0.03)({2\pi\! Rh})=0.1\pi\! R^{2}+0.06\pi\! Rh$$

By substituting for $h$, we can express $C$ as a function of $R$. $$C=C( R) =0.1\pi\! R^{2}+({0.06\pi\! R})\left( {\frac{1000}{\pi\! R^{2}}} \right)\! =0.1\pi\! R^{2}+\frac{60}{R}$$

This is the function to be minimized. The domain of $C$ is $\left\{ R|R>0\right\}$.

To find the minimum cost, we differentiate $C$ with respect to $R$. $${C}' (R)=0.2\pi\! R-\frac{60}{R^{2}}=\frac{0.2\pi\! R^{3}-60}{R^{2}}$$

Solve $C' (R)=0$ to find the critical numbers. $$\begin{eqnarray*} 0.2\pi\! R^{3}-60 &=&0 \\[1pt] R^{3} &=&\frac{300}{\pi } \\[1pt] R &=&\sqrt[3]{\frac{300}{\pi }}\approx 4.571\,\ {\rm{cm}} \end{eqnarray*}$$

Now we find $C'' (x)$ and use the Second Derivative Test. $$\begin{eqnarray*} {C}'' (R) &=&0.2\pi +\frac{120}{R^{3}} \\[3pt] {C}'' \left( \sqrt[3]{\frac{300}{\pi }}\right) &=&0.2\pi +\frac{ 120\pi }{300}=0.6\pi >0 \end{eqnarray*}$$

$C$ has a local minimum at $\sqrt[3]{\dfrac{300}{\pi }}$. Since $C'' (R)>0$ for all $R$ in the domain, the graph of $C$ is always concave up, and the local minimum value is the absolute minimum value. The radius of the container that minimizes the cost is $R\approx 4.571\ {\rm{cm}}.$ The height of the container that minimizes the cost of the material is $$h=\frac{1000}{\pi\! R^{2}}\approx \frac{1000}{20.892\pi }\approx 15.236\,\ {\rm{cm}}$$

The minimum cost of the container is $$C\left( \sqrt[3]{\dfrac{300}{\pi }} \right) =0.1\pi \left( \sqrt[3]{\dfrac{300}{\pi }}\right) ^{2}+\dfrac{60}{ \sqrt[3]{\dfrac{300}{\pi }}}\approx \$19.69.$$