Question

Cylindrical soup cans are to be manufactured to contain a given volume V . No waste is involved in cutting the material for the vertical side of each can, but each top and bottom which are circles of radius r, are cut from a square that measures 2r units on each side. Thus the material used to manufacture each soup can has an area of A = 2πrh + 8r2 square units.

(a) How much material is wasted in making each soup can?

(b) Find the ratio of the height to diameter for the most economical can (i.e. requiring the least amount of material for manufacture.)

(c) Use either the first or second derivative test to verify that you have min-imized the amount of material used for making each can.

Answer 1

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

(a)

$$A=2\pi rh+8r^2$$

$$A(r)=2\pi r\cdot\dfrac{V}{\pi r^2}+8r^2$$

$$A(r)=\dfrac{2V}{r}+8r^2, r>0$$

is wasted in making each soup can

(b)

$$\dfrac{dA}{dr}=-\dfrac{2V}{r^2 }+16r$$

$$\dfrac{dA}{dr}=0=>-\dfrac{2V}{r^2}+16r=0$$

$$r=\dfrac{\sqrt[3]{V}}{2}$$

is the critical number. Then

$$h=\dfrac{V}{\pi ({\sqrt[3]{V} \over 2})^2}=\dfrac{4\sqrt[3]{V}}{\pi}$$

The ratio of the height to diameter for the most economical can

$$\dfrac{h}{2r}=\dfrac{\dfrac{4\sqrt[3]{V}}{\pi}}{2\cdot \dfrac{\sqrt[3]{V}}{2}}=\dfrac{4}{\pi}$$

$$\dfrac{h}{2r}=\dfrac{4}{\pi}$$

(c)

The first derivative test

If $0<r<\dfrac{\sqrt[3]{V}}{2},$ then $\dfrac{dA}{dr}<0, A$ decreases.

If $r>\dfrac{\sqrt[3]{V}}{2},$ then $\dfrac{dA}{dr}>0, A$ increases.

The function $A$ has a local minimum at $r=\dfrac{\sqrt[3]{V}}{2}.$

Since the function $A$ has the only extremum, then the function $A$ has the absolute minimum at $r=\dfrac{\sqrt[3]{V}}{2}.$

Hence the most economical can will be if

$$r=\dfrac{\sqrt[3]{V}}{2},\dfrac{h}{2r}=\dfrac{4}{\pi}$$