Subsections

• Purpose
• Background
  • Locating Global Extrema
  • Maple Commands
  
  
• Exercises

MA 1024 Lab: Local and Global Extrema

Purpose

The purpose of this lab is to acquaint you with techniques for finding global extreme values of functions of two variables.

Background

Many applications of calculus involve finding the maximum and minimum values of functions. For example, suppose that there is a network of electrical power generating stations, each with its own cost for producing power, with the cost per unit of power at each station changing with the amount of power it generates. An important problem for the network operators is to determine how much power each station should generate to minimize the total cost of generating a given amount of power.

A crucial first step in solving such problems is being able to find and classify local extreme values of a function. What we mean by a function $f$ having a local extreme value at a point $\mathbf{p}_0$ is that for values of $\mathbf{p}$ near $\mathbf{p}_0$, $f(\mathbf{p}_0) \geq f(\mathbf{p})$ for a local maximum and $f(\mathbf{p}_0) \leq f(\mathbf{p})$ for a local minimum.

Locating Global Extrema

In one-dimensional calculus, the absolute or global extreme values of a function occur either at a point where the derivative is zero, a boundary point, or where the derivative fails to exist. The situation for a function of two variables is very similar, but the problem is much more difficult because the boundary now consists of curves instead of just endpoints of intervals. For example, suppose that we wanted to find the global extreme values of a function $f(x,y)$ on the rectangle $S = \{(x,y): a \leq x \leq b \mbox{ and } c \leq y \leq d\}$. The boundary of this rectangle consists of the four line segments given below.

$$\begin{array}{lcrc} y & = & c, & a \leq x \leq b \\ y & = &... ... & c \leq y \leq d \\ x & = & b, & c \leq y \leq d \end{array}$$

The basic theorem on the existence of global maximum and minimum values is the following.

Theorem 1   Suppose $f(x,y)$ is continuous on a closed, bounded set $S$, then $f(x,y)$ attains its absolute maximum value at some point $(x_0,y_0)$ in $S$ and absolute minimum value at some point $x_1,y_1)$ in $S$.

This theorem only says that the extrema exist, but doesn't help at all in finding them. However, we know that the global extrema occur either at local extrema, on the boundary of the region, or at points where one or the other partial derivative fails to exist. For example, to find the extreme values of a function $f(x,y)$ on the rectangle given above, you would first have to find the interior critical points and then find the extreme values for the four one-dimensional functions

$$\begin{array}{lccc} g_1(x) & = & f(x,c), & a \leq x \leq b \... ...y \leq d \\ h_2(y) & = & f(b,y), & c \leq y \leq d \end{array}$$

Maple Commands

In order to find the absolute (or global) extrema given the paraboloid $f(x,y)=2+x^2+y^2$ whose domain is the rectangle made of the four lines $y = -1; y = 3; x = -1; x = 3$, first enter the three-dimensional function.

>f:=(x,y)->2+x^2+y^2;

Keep your work organized; you may want to work your way through dimensions. Starting with the three dimensional function find the critical points by setting both partials equal to zero; you can do this in one command line. (Notice that there are no undefined points)

>solve({diff(f(x,y),x)=0,diff(f(x,y),y)=0},{x,y});

Next find the critical points along the two-dimensional domain which is the rectangular boundary.The trick is to replace one of the variables with part of the boundary.Since there are four 2-d functions (or sides) this will be done four times.

>solve(diff(f(-1,y),y)=0);
>solve(diff(f(3,y),y)=0);
>solve(diff(f(x,-1),x)=0);
>solve(diff(f(x,3),x)=0);

This gives us the four points $(-1, 0), (3, 0), (0, -1)$, and $(0, 3)$. Now the one-dimensional domain is simply the corners: $(-1, -1), (-1, 3), (3, -1)$, and $(3, 3)$. Now that you have all possible points listed you simply need to plug them all into the original function to find the nine z-values. One point plugged into $f$ is shown below. The highest and lowest will be the absolute maximum and absolute minimum.

>evalf(f(-1,0))

Exercises

1. Find the global extrema of
   
   $$f(x,y) = y^2-6y\sqrt{x}+3xy$$
   
   
   over the region bounded by the rectangle $0 \leq x \leq 3$, $-1 \leq y \leq 2$.

2. Find the global extrema of
   
   $$f(x,y) = x-\frac{x^3}{8}-\frac{y^2}{9}$$
   
   
   over the elliptical region $R$ bounded by $$\frac{x^2}{4}+y^2 \leq 1$$.