VIDEO: Solving simple extreme value problems

Modeling the extreme value problems

• First you have to set up a functional equation f, which is dependent on a parameter, usually x is used. x denotes the variable and unknown quantity that must be selected so that a maximum or minimum result for the extreme value problem is achieved in the end.
• x can be B. stand for the length of a table or the weight of a brick.
• You then have z. B. a function of the form f (x) = 2x³-4x + 3 found.
• But it can also be that the function is dependent on two or more variables in the first step, e.g. B. f (x, y) = 5x²-2xy + 3y-6.
• Now you have to find a constraint that specifies one variable as a function of the other variable. Applies e.g. B. y = 2x + 2, then you can insert this y into the functional equation and you now get a simple functional equation that only depends on x. In this example, after multiplying and combining, this would be: f (x) = 5x²-2x (2x + 2) +3 (2x + 2) -6 = x²+ 2x + 6.


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• This example is examined further below.

Simple differentiation - that's how it works

• Once you have found the function equation that models your extreme value problem, all you need to do is find the special value for x that minimizes or maximizes your function.
• To do this, you have to take the first derivative of the function with respect to x. For this you might need the product, quotient or chain rule, depending on the difficulty of the function equation. If you are no longer familiar with this from school, you can find it in simple derivation rules in popular formulas or books.
• In our example we now get the derivative function f '(x) = 2x + 2.
• You have to know that there can only be an extreme point where the condition f '(x) = 0 is fulfilled.
• So in the next step you have to set the derivative equal to 0. In this example this would be 0 = f '(x) = 2x + 2 <=> 2x = -2 <=> x = -1.
• At the point x = -1 there is therefore a candidate for an extreme point.
• Of course, there could be multiple candidates for your extreme value problems. These must also be checked individually in the next step. In this simple example there is only one candidate.

Simple differentiation successful - what now?

• In order to find out whether there are simple extreme points at the determined points, the second derivative must be formed.
• There are three possibilities: f '' (x) <0 applies, here there is a local maximum. Or: f '' (x)> 0 applies, here there is a local minimum. Or: f '' (x) = 0, there is no extreme point here (it is a so-called saddle point).
• In the simple example discussed here, the second derivative must be examined at the point x = -1. First of all, f '' (x) = 2. So also f '' (- 1) = 2.
• Because of f '' (- 1)> 0 there is a local minimum at the point x = -1.
• If you have found other candidates for your extreme value problems, you should now also check for each candidate whether there is an extreme point and what type it is.

As you can see, it is really easy to find a solution to most extreme value problems. The greatest difficulty lies only in setting up the correct functional equation for the respective extreme value problem.