Here's a problem from Amanda's most recent calculus quiz. A print shop is to make a poster with a four-inch margin at the top and bottom, a two-inch margin on the sides, and a print area of 50 square inches. What are the dimensions of the smallest poster--least area--that can be made.

Well, if the horizontal print area is x, and the vertical print area y, then xy = 50 and y = 50/x. The area of the entire poster might then be expressed by the equation (x + 4) (50/x + 8). Expand this equation and you get 50 + 8x + 200/x + 32. Differentiating, we arrive at 8 - 200/x ^{2}. So an extreme value exists where 8 = 200/x ^{2}, which would be when x = 5. And when x = 5, y = 10. Alack, this is where I quit when working the problem, which did not ask anything about the print area but rather about the dimensions of the whole poster. That would be (5 + 4) x (10 + 8), or 9 x 18 inches.

Martin Gardner, in *Calculus Made Easy*, is fond of similar problems involving various shapes or solids inside of bigger ones--for example, find the volume of the largest cylinder inscribed within a cone the height of which is the same as the radius of its base. It would help to draw a picture but I am not clever enough for that--even if it is possible. You will have to imagine a cone--the often orange thing that kids use to mark a goal for their pick-up soccer games, or that the highway department uses to keep cars off a patch of road they are working on. The cone the problem describes has a height equal to the radius of its base and it also has a cylinder--a solid shaped like a can of anything at the grocery store--inside of it. If the cylinder is tall, its radius will be small; and as the radius gets bigger and bigger, finally approaching that of the cone, its height shrinks and the shape becomes "squat." Its radius can't be greater than that of the cone or else it won't "fit."

The volume of a cylinder is the procuct of pi, its height, and its radius squared. The one we are talking about is limited in the respect that it must be inscribed within the cone, which is to say that the sum of its height and radius is equal to the radius (which is the same as the height) of the cone. The state of affairs suggests the following algebraic expression

h + r = R

where h refers to the height of the cylinder, r to the radius of the cylinder, and R to the radius of the cone. It follows that

h = R - r

in which case the volume of the cylinder may be rewritten thus to eliminate one variable:

A = (pi) r

^{2}(R - r) = (pi) R r^{2}- (pi) r^{3}.

Now we differentiate and set the derivative equal to zero to determine the extreme value:

A' = 2 (pi) R r - 3 (pi) r

^{2}

2 (pi) R r = 3 (pi) r^{2}

2 R = 3 r

r = 2R/3

To maximize the volume of this inscribed cylinder, make its radius two-thirds that of the cone.

It may seem strange that the solutions to both problems are so regular and involve 2/3 or the ratio 2-to-1 instead of some more complicated fraction. But if you work a lot of these problems you begin to notice that this is almost always the case. The poster with the smallest area has one dimension exactly twice that of the other. The cylinder with the greatest volume has a radius exactly two-thirds that of the cone in which it is inscribed. You can verify, if you are a dweeb, that a cylinder of a given volume will have the maximum surface area if its height is exactly twice the length of its radius.

## Comments