Answer to Question #272638 in Calculus for Zarina

Question

A cylindrical can is to be constructed to hold a fixed volume of liquid. The cost of the

material used for the top and bottom of the can is 3 cents per square inch, and the cost of the

material used for the curved side is 2 cents per square inch. Use calculus to derive a simple

relationship between the radius and height of the can that is the least costly to construct.

Answer 1

Let $r=$ rafius of the cylinder, $h=$ its height.

$$V=\pi r^2h=>h=\dfrac{V}{\pi r^2}$$

The cost of the material is

$$C=0.03(2\pi r^2)+0.02(2\pi r h)$$

Substitute

$$C=C(r)=0.03(2\pi r^2)+0.02(2\pi r (\dfrac{V}{\pi r^2}))$$

$$=0.06\pi r^2+\dfrac{0.04V}{r}$$

Find the first derivative

$$C'(r)=0.12\pi r-\dfrac{0.04V}{r^2}$$

Find the critical number(s)

$$C'(r)=0=>0.12\pi r-\dfrac{0.04V}{r^2}=0$$

$$r=\sqrt[3]{\dfrac{V}{3\pi}}$$

If $0<r<\sqrt[3]{\dfrac{V}{3\pi}}, C'(r)<0, C(r)$ decreases.

If $r>\sqrt[3]{\dfrac{V}{3\pi}}, C'(r)>0, C(r)$ increases.

The function $C(r)$ has a local minimum at $r=\sqrt[3]{\dfrac{V}{3\pi}} .$

Since the function $C(r)$ has the only extremum. the function $C$has the absolute minimum at $r=\sqrt[3]{\dfrac{V}{3\pi}} .$

$$h=\dfrac{V}{\pi r^2}=\sqrt[3]{\dfrac{9V}{\pi}}$$