Third Example: Modeling (1.4) and

Practical Optimization (3.5)

An agent wants to maximize (or minimize) a certain variable, which we call the target variable. He or she does this by choosing the value for some so-called choice variable. There may be more variables involved, we call these the other variables. Consider the following example:

A cone is made from a circular piece of paper (of radius 10) by cutting off a sector and glueing together the blue edges AB and AC. Find the maximum volume of such a cone. Note: The volume of a cone of height h and radius r for the base circle equals

We assume that x is the fraction of new area divided by old area. It is also the fraction of the (red= part of the perimeter divided by perimeter before. x is our choice variable.

The dimensions of the cone, its height h and the radius r of the base circle, are the "other" variables.

The target variable is the volume V of the resulting cone.

We need to express the target variable in terms of the choice and other variables.

in our example:

Now, since the target variable should become a function of the choice variable alone, every other variable occurring in the above equation must be expressed in terms of the choice variable and replaced. Usually we need as many equations between the variables as we have "other" (non-target and non-choice) variables. Ideally each such equation is just an equation between such an "other" variable and the choice variable, but it could be more complicated.

First we express h in terms of r by looking on the sketch of the cone to the right. Using Pythagoras' Theorem, we get (since the blue line is just our glued edge AC and has therefore length 10)

and thus .

Substituting this into our V formula gives

Since the perimeter before was , it is now . For the resulting smaller circle which forms the base of the cone, this implies that its radius r equals 10x. We substitute this again to get the target variable as a function of the choice variable alone, as desired: .

Now we have a functional relation between choice and target variable. We differentiate this function with respect to choice variable, set it equal to 0, find the critical points, and so on.

Maybe we graph the function first. It seems that x=0.85 would result in a maximum volume of about 400.

Differentiating the function V(x), we get or .

We set V'(x)=0, multiply the resulting equation by the radical to get . The solutions are x=0, and the solutions of , where the positive one, is the solution of our problem.

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Erich Prisner, November 2004