The Extreme Value Theorem

The extreme value theorem is a theorem in several-variable calculus that we can use to find extreme values and simultaneously prove rigorously that they are extreme values. To state it requires some definitions concerning subsets of R².

Open Disks

Suppose P=(a,b) is a point in R² and r is a positive real number. The open disk of radius r centered at P is the set of all (x,y) in R² such that (x-a)² + (y-b)² < r². In symbols, the open disk D of radius r centered at P is the set D:={(x,y)∈R²: (x-a)² + (y-b)² < r²}.

Boundary and Interior Points

Suppose S is a subset of R². A point P in R² is called a boundary point of S if every open disk centered at P contains both points that are in S and points that are not in S.

For example, suppose K is the set of all points (x,y) such that x² + y² ≤ 1. Then the point (1,0) is in the boundary of K. However, the point (0,0) is not in the boundary of K because there are open disks centered at (0,0) that are totally contained in K. Similarly the point (2,0) is not in the boundary of K because there are open disks centered at (0,0) that do not intersect K at all.

A point P in a subset S of R² is said to be in the interior of S if P does not lie in the boundary of S.

For example, the point (0,0) lies in the interior of the set K above, but (1,0) does not as it is on the boundary and (2,0) does not because it is not in K.

Closed Subsets

A subset S of R² is said to be closed if it contains all of its boundary points.

For example, the closed disk K:={(x,y)∈R²:x² + y² ≤ 1} is a closed subset. However, the open disk D:={(x,y):x² + y² < 1} is not because it does not contain the point (1,0) which is a boundary point of D. The half-open square T:={(x,y): 0≤ x < 1, 0≤ y< 1} is also not closed because it does not contain (1,1) which is a boundary point of T.

Bounded Sets

A subset S of R² is bounded if it can be contained inside a disk. For example, the set T:={(x,y): 0≤ x < 1, 0≤ y< 1} is bounded because it can be contained inside the open disk of radius 2 centered at 0. (In fact, it can even be contained inside the open disk of radius ${\sqrt {2}}$ centered at 0.) The set L:={(x,y): 0<x<1} is not bounded. It looks like an infinite strip running along the y-axis.

The Extreme Value Theorem

Theorem. If f:S→R is a continuous function with S a closed and bounded subset of R², then f attains an absolute maximum and an absolute minimum value on S.

Hypotheses and Counterexamples

The hypotheses continuous, closed and bounded are all necessary for the conclusion of the theorem to hold. For example, suppose D is the open disk of radius 1 centered at the origin (which is not a closed subset of R²) and f is the function f(x,y)=x The function f does not attain a maximum on D even though f is continuous and D is bounded. There are similar examples showing that the other hypotheses are necessary.

Applications of the Extreme Value Theorem (EVT)

Method for computing extreme values

The theorem gives us a method (outlined in our text) for finding extreme values of functions of two variables on a closed and bounded subset S of R².

Evaluate f at the critical points of f in the interior of S.
Compute the extreme values of f along the boundary of S.
The largest values obtained in steps 1 and 2 is the absolute maximum value of f on S; the smallest is the absolute minimum.

The reason this works is that, since S is closed and bounded, it must have some maximum value f(P). If P is in the interior of S, then f(P) is a local maximum of f. Therefore P must be a critical point of f and we will compute its value in step 1. If P lies on the boundary, we will compute its value in step 2.

An Example

Find the absolute minimum and absolute maximum of the function f(x,y)=2x² + xy + y² + y on the square S:={(x,y):-1≤ x≤ 1, -1≤y ≤ 1}.

Solution. The first step is to find the critical points of f on the interior of the square S. So we compute

$\nabla f(x,y)=\langle 4x+y,x+2y+1\rangle$

The only solution of the equation ∇ f(x,y)=0 is (x,y)=(1/7, -4/7). This point lies in the interior of S, and it is the only critical point of S there. We have f(1/7,-4/7)=-2/7.

The boundary. Now we need to compute the extrema of f on the boundary of S. The boundary consists of 4 line segments: the segment A from (1,-1) to (1,1), the segment B from (1,1) to (-1,1), the segment C from (-1,1) to (-1,-1), and finally the segment D from (-1,-1) to (1,-1). We need to find the extreme values of f on A, B, C and D. But before we do we record the values of f at the corners of the square: f(1,-1)=1, f(1,1)=5, and f(-1,1)=f(-1,-1)=3.

Now we look at the boundary intervals.

On A, we have x=1. So we need to find the extreme values of the function f(1,y)=y² + 2y + 2 as y goes from -1 to 1. We have (d/dy) f(1,y)=2y+2 so y=-1 is the only critical point of this function. We have f(1,-1)=1. It follows that the maximum value of f on A is f(1,1)=5 and the minimum is f(1,-1)=1.

On B, we have y=1. So we need to find the extreme values of the function f(x,1)=2x² + x + 2. We have (d/dx) f(x,1)=4x+1. The only critical point is x=-1/4 and we have f(-1/4,1)=15/8. It follows that the maximum of f on B is 5 and the minimum is 15/8.

Similarly, the minimum of f on C is 2=f(-1,0) and the maximum is 3. Finally, on D we have y=-1. So we look at the function f(x,-1)=2x² -x. We have (d/dx) 2x² -x = 4x-1 with x=1/4 as the only critical point. We have f(1/4,-1)=-1/8. So -1/8 is the minimum of f on D and f(-1,-1)=3 is the maximum.

The maximum on the boundary is 5 and the minimum is -1/8.

Final Answer. Putting it all together we have f(1,1)=5 is the maxmimum and f(1/7,-4/7)= -2/7 is the minimum since it is smaller than -1/8. You can check our work by looking at the following table of values.

P	(1/7,-4/7)	(1,-1)	(1,1)	(-1,1)	(-1,-1)	(-1/4,1)	(-1,0)	(1/4,-1)
f(P)	-2/7	1	5	3	3	15/8	2	-1/8