Q. 57 - Calculus

Question

The cost of the material for the top and bottom of a cylindrical can is $5$ cents per square inch. The material for the rest of the can cost only $2$ cents per square inch. If the can must hold $400$ cubic inches of liquid, what is the cheapest way to make the can? What is the most expensive way?

Step-by-Step Solution

Answer

Ans:  To get the minimum cost, the can should be constructed as $2.94$ inches high with a radius of $14.73$ inches.  There is no global maximum.

Step 1. Given information.

given,

     The cost of the material for the top and the bottom of a cylindrical can is 5 cents per square inch.
 The cost of the rest of the material is 2 cents per square inch.
 The cylindrical can holds 400 cubic feet of oil.

Step 2. The objective is to find how should the oil drums can be constructed so that they use as little metal as possıble.

The surface area of the cylinder is as follows.

   $$\begin{gathered} S=2{\mathrm{\pi r}}^2+2\mathrm{\pi rh} \\ S=10{\mathrm{\pi r}}^2+4\mathrm{\pi rh} \end{gathered}$$

The volume of the box is as follows,

    $$\begin{gathered} V={\mathrm{\pi r}}^2\mathrm{h} \\ 400={\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{h}=\frac{400}{{\mathrm{\pi r}}^2} \end{gathered}$$

Step 3. The derivative of the surface area is calculated below.

 $\begin{array}{r}S=10\pi r^2+4\pi r\left(\frac{400}{\pi r^2}\right)\\ S=10\pi r^2+\frac{1600}{r}\\ 20\pi r^3-1600=0\\ 20\pi r^3=1600\\ \pi r^3=80\\ r=\sqrt[3]{\frac{80}{\pi }}\\ =2.94\end{array}$

Step 4. The height of the cylinder is calculated below.

  $\begin{array}{r}h=\frac{400}{\pi r^2} \\ =\frac{40}{\pi \times (2.94{)}^2}\\ =14.73 \end{array}$

Therefore, to get the minimum cost, the can should be constructed as $2.94$ inches high with a radius of $14.73$ inches.  There is no global maximum