Previously, we solved an optimization problem using the Simplex Method. That method works when both the objective function and the constraints are linear. That is, when we have a linear program (LP).
What if the function is nonlinear? For a one-variable function on a closed interval, a standard calculus approach is to find its critical points, then compare the function’s values at those points and at the interval’s endpoints.
The second derivative can help us classify critical points:
- If f''(x)<0, the function f(x) has a local maximum at x
- If f''(x)>0, the function f(x) has a local minimum at x
- If f''(x)=0, the test is inconclusive
Let’s work through an example. Suppose f is defined on [-2,3] by:
First, take the derivative and set it equal to zero:
Factoring gives:
Hence, we have two cases for the equation to equal zero:
Case 1 implies:
Case 2 implies:
So the critical points are:
Now find the second derivative:
At the critical points:
To find the absolute maximum and minimum on the closed interval, we compare the function’s values at the critical points and at both endpoints:
The greatest value is 19, at x=-1. The least value is -45, at x=3.
So, on [-2,3], the absolute maximum is f(-1)=19 and the absolute minimum is f(3)=-45.
The other interior extrema are local:
A minimum at x=1 and a maximum at x=2.
Graph:
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