Constrained Extrema

Sometimes, the absolute extrema (maximum/minimum) of a function may be restricted with certain constraints, especially in linear programs.

Existence of Absolute Extrema §

The extreme value theorem guarantees the existence of absolute extrema, provided the domain of the function is both closed and bounded.

• For example, the theorem does not hold if the domain is ${\mathbb{R}}^k$, because that is unbounded. Mind you, we may still be able to find absolute extrema, they just cannot be guaranteed by the theorem.

Finding Absolute Extrema §

Let $f:D\to \mathbb{R}$, where $D$ is closed and bounded in ${\mathbb{R}}^n$

1. Find the values of $f$ on critical points in $D$
2. Find the values of $f$ on boundary points in $D$
3. Compare them to find extrema.

Examples §

E1: Triangle Constraint

Find the absolute extrema of the function $f(x,y)=xy-x-y+3$ on the triangle with vertices $(0,0),(2,0),(0,4)$: $f$ is a polynomial, and hence is infinitely differentiable. So, it only has stationary points, given by $\nabla f(x,y)=\vec{0}$:

$$\nabla f(x,y)=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)=(y-1,x-1)$$

A stationary point is $(1,1)$. We can also use the second partial derivative test to see if $(1,1)$ is worth checking (if it is a saddle point, it isn’t!). The Hessian returns:

$$\det ({\mathbf{H}}_f)=f_{xx}{f}_{yy}-(f_{xy}{)}^2=0-1=-1$$

So $(1,1)$ is a saddle point. We can ignore it. Let’s check the boundary points next:

• $f(0,0)=3$
• $f(2,0)=1$
• $f(0,4)=-1$

Thus the absolute maximum is 3 (at the point (0,0)) and the absolute minimum is -1 (at the point (0,4))