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.
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Geometry formulas are discussed in Appendix A.2, p. A-15.
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. \[ \begin{eqnarray*} {C}'' (R) &=&0.2\pi +\frac{120}{R^{3}} \\[3pt] {C}'' \left( \sqrt[3]{\frac{300}{\pi }}\right) &=&0.2\pi +\frac{ 120\pi }{300}=0.6\pi >0 \end{eqnarray*} \]
If the costs of the materials for the top, bottom, and lateral surfaces of a cylindrical container are all equal, then the minimum total cost occurs when the surface area is minimum. It can be shown (see Problem 39) that for any fixed volume, the minimum surface area of a cylindrical container is obtained when the height equals twice the radius.
\(C\) 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 \[ C\left( \sqrt[3]{\dfrac{300}{\pi }} \right) =0.1\pi \left( \sqrt[3]{\dfrac{300}{\pi }}\right) ^{2}+\dfrac{60}{ \sqrt[3]{\dfrac{300}{\pi }}}\approx \$19.69. \]