Second 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:

Two vertical poles, AB and CD, both 2m high and standing 3m apart, are connected by a rope, which is also anchored on the ground at point E. For which choice of E is the length of the rope minimized?

We take the same picture and label everything we don't know yet.

Choice variable: x

target variable: f=z+w

other variables: y, z, w

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

in our example: f=z+w

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.

We use the Theorem of Pythagoras in the two right-angled triangels:

, which solved for z yields , and

, which we reformulate into .

We substitute z and w by these expressions to get

The only remaining non-choice variable on the right side is y. All what remains to be done is express y also by x. We use the information that and get . Plugging this into the equation above yields =

The first equation is nice: just between choice variable x and other variable z.

The second equation however relates the two "other" variables y and w.

The third equation relates "other" variable y and choice variable x.

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.

We have the function

It's graph is shown to the right, and it seems obvious that the minimum is achieved for x=1.5.

We diferentiate to get

The equation f'(x)=0 we solve by multiplying the equation by both radicals (to get rid of the fractions), and then we solve the resulting radical equation using College Algebra techniques. The solution is x=1.5.

Look for another example

Erich Prisner, November 2004