Problem

If the cost of material to make the can in Example 1 is 5 cents per square inch for the to...

If the cost of material to make the can in Example 1 is 5 cents per square inch for the top and bottom and 3 cents per square inch for the sides, what dimensions should be used to minimize the cost of making the can? [The answer is not the same as in Example 1.]

EXAMPLE 1

A cylindrical can of volume 58 cubic inches (approximately 1 quart) is to be designed. For convenient handling, it must be at least 1 inch high and 2 inches in diameter. What dimensions will use the least amount of material?

SOLUTION

We can construct a can by rolling a rectangular sheet of metal into a tube and then attaching the top and bottom, as shown in Figure 1. The surface area of the can (which determines the amount of material) is

When the sheet is rolled into a tube, the width c of the sheet is the circumference of the end of the can, so that c = 2πr, and hence,

Surface area = ch + 2πr2 = 2πrh + 2πr2

The volume of a cylinder of radius r and height h is πr2h. Since the can is to have volume 58 cubic inches, we have

Therefore,

Note that r must be greater than 1 (since the diameter 2r must be at least 2). Furthermore, r cannot be more than 5 (if r > 5 and h ≥ 1, then the volume πr2h would be at least π · 25 · 1, which is greater than 58).

The situation can be represented by the graph of the equation

y = 116/x +2πx2

as in Figure 2. The x-coordinate of each point represents a possible radius, and the y-coordinate represents the surface area of the corresponding can. We must find the point with the smallest y-coordinate, that is, the lowest point on the graph. A graphical minimum finder (Figure 2) shows that the coordinates of this point are approximately (2.09773, 82.946845). If the radius is 2.09773, then the height is 58/(π2.097732) ≈ 4.1955. As a practical matter, it would probably be best to round to one decimal place and construct a can of radius 2.1 and height 4.2 inches.

Figure 1