Q. 58 - Calculus

Question

Consider the can-making situation in the previous exercise, but suppose that the cans are made with open tops.

If each can must hold 400 cubic inches of liquid, what is the cheapest way to make the cans? What is the most expensive way?

Step-by-Step Solution

Answer

The value of the radius is 3.706 and height is 9.266.

Step 1.Given Information

Given the cost of material for bottom of an cylindrical can is 5cents per square inches.

The cost of rest of the material is 2cents per square inches.

The cylindrical can holds 400 cubic per feet of oil.

Step 2.The formula

The surface area of the cylinder is as follows:

$$\begin{gathered} s={\mathrm{\pi r}}^2+2\mathrm{\pi rh} \\ \mathrm{s}=5{\mathrm{\pi r}}^2+4\mathrm{\pi rh} . \end{gathered}$$

The volume of the box is as:

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

Step 3.The radius of the cylinder

Derivative of the surface area is calculated as:

$$\begin{gathered} s=5{\mathrm{\pi r}}^2+4\mathrm{\pi r}\left(\frac{400}{{\mathrm{\pi r}}^2}\right) \\ \mathrm{s}=5{\mathrm{\pi r}}^2+\frac{1600}{\mathrm{r}} \\ {\mathrm{s}}^{\text{'}}=10\mathrm{\pi r}-\frac{1600}{{\mathrm{r}}^2} . \end{gathered}$$

$$\begin{gathered} s^{\text{'}}=0 \\ 10\mathrm{\pi r}-\frac{1600}{{\mathrm{r}}^2}=0 \\ 10{\mathrm{\pi r}}^3=1600 \\ \mathrm{r}=\sqrt[3]{\frac{1600}{\mathrm{\pi }}} \\ =3.706 . \end{gathered}$$

Step 3.The height

The height of the cylinder is calculated as:

$$\begin{gathered} h=\frac{400}{{\mathrm{\pi r}}^2} \\ =\frac{400}{\mathrm{\pi }{\left(3.706\right)}^2} \\ =9.266 . \end{gathered}$$

The value of the radius is 3.706 and height is 9.266.