In this set of notes, we are going to learn how to find the extreme values for functions of two variables. To start the topic off, let us state a fact about continuous functions.

FACT:

Continuous functions defined on closed bounded regions in the xy-plane take on absolute maximum and minimum values on these domains.

Remember how we found extrema in calculus I? We first found the first derivative, then sat it equal to zero to find the critical points. Then we found the second derivative, plugged in the critical point, and determined if it was concave up or concave down. We will do the same thing now, except we are dealing with partial derivatives.

Do you remember what we were looking for when we sat the first derivative equal to zero in calculus I? We were looking for points where the graph had a horizontal tangent line. Then we looked for local maxima, local minima, and points of inflection.

Now, for a function z = f (x, y), we will look for points where the surface has a horizontal tangent plane. At these points, we will look for local maxima, local minima, and saddle points.

DEFINITION:

Let f (x, y) be defined on a region R containing a point (a, b). Then

1.

f (a, b) is a local maximum value of f if f (a, b) ³ f (x, y) for all domain points (x, y) in an open disk centered at (a, b).

2.

f (a, b) is a local minimum value of f if f (a, b) £ f (x, y) for all domain points (x, y) in an open disk centered at (a, b).

You can think of a local maxima as being the mountain peaks on the surface z = f (x, y), and the local minima as the valleys. (See figure 1)

figure 1

FACT:

If f (x, y) has a local maximum or minimum value at an interior point (a, b) of its domain, and if the partial derivatives exist there, then f _x (a, b) = 0 and f _y (a, b) = 0.

As in the single variable case, the only places a function f (x, y) can ever have an extreme value are (1) interior points where f _x = f _y = 0. (2) Interior points where one or both f _x and f_y do not exist, and (3) boundary points of the function's domain.

DEFINITION:

An interior point of the domain of a function f (x, y) where f _x and f _y are zero or where one or both f _x and f _y do not exist is a critical point of f.

So we can safely assume that the only points where a function f (x, y) can assume extreme values are the critical points and boundary points. We should always keep in mind that not every critical point is a local extrema, and that a differentiable function of two variables might have a saddle point.

DEFINITION:

A differentiable function f (x, y) has a saddle point at a critical point (a, b) if every open disk centered at (a, b) there are domain points (x, y) where f (x, y) > f (a, b) and domain points (x, y) where f (x, y) < f (a, b). The corresponding point (a, b, f (a, b)) on the surface z = f (x, y) is called a saddle point of the surface.

A good example of a function that has a saddle point is a hyperbolic paraboloid. If you need to see one, check out supplemental notes 6. Now let us determine the critical points of a function of two variables.

EXAMPLE 1a:

Find all the critical points of f (x, y) = 2xy - 5x ² - 2y ² + 4x + 4y - 4.

SOLUTION:

The only critical point is (2/3, 4/3).

How do we determine if this critical point is a local extreme point? We will use the second derivative test for local extreme values.

Suppose f (x, y) and its first and second partial derivatives are continuous throughout a disk centered at (a, b) and that f _x (a, b) = f _y (a, b) = 0. Then

i.

f has a local maximum at (a, b) if f _{x x} < 0 and f _{x x} f _{y y} - f _{x y}² > 0 at (a, b).

ii.

f has a local minimum at (a, b) if f _{x x} > 0 and f _{x x} f _{y y} - f _{x y}² > 0 at (a, b).

iii.

f has a saddle point at (a, b) if f _{x x} f _{y y} - f _{x y}² < 0 at (a, b).

iv.

The test is inconclusive at (a, b) if f _{x x} f _{y y} - f _{x y}² = 0 at (a, b). In this case, we must find some other way to determine the behavior of f at (a, b).

EXAMPLE 1b:

Determine if the critical point found in 1a is a local maximum, minimum, or saddle point.

SOLUTION:

(2/3, 4/3) is a local minimum and f (2/3, 4/3) = 0.

 EXAMPLE 2:

Find all local maxima, local minima, and saddle points of the function f(x,y) = x ² - 4xy + y ² + 6y + 2.  

SOLUTION:

The only critical point for this function is the point (2, 1).

Now to find the second partial derivatives.

The point (2, 1) is a saddle point, and the value of the function at this point is f = 5.

EXAMPLE 3:

Find all local maxima, local minima, and saddle points of the function f(x,y) = 2x ² + 3xy + 4y ² - 5x + 2y.

SOLUTION:

This function has only one critical point, and it is (2, -1).

(2, -1) is a local minimum, and the value of the function evaluated at this point is f = -6.

EXAMPLE 4:

Find all local maxima, local minima, and saddle points for the function f(x,y) = x ³ + 3xy + y ³.

SOLUTION:

The critical points for this function are (0, 0) and (-1, -1).

The point (0, 0) is a saddle point, and f (0, 0) = 0.

The point (-1, -1) is a local maximum, and f (-1, -1) = 1.

EXAMPLE 5:

Find all local maxima, local minima, and saddle points of the function f(x,y) = x ⁴ + y ⁴ + 4xy.

SOLUTION:

The critical points for this function are (0, 0), (1, -1), and (-1, 1).

The point (0, 0) is a saddle point, and f (0, 0) = 0.

The point (1, -1) is a local minimum, and f (1, -1) = -2.

The point (-1, 1) is a local minimum, and f (-1, 1) = -2.

Here are the steps you should follow to find the absolute extrema on a closed bounded region.

STEP 1:

List interior points of R where f _x = f _y = 0 or one or both f _x and f _y fail to exist. Evaluate f (x, y) at these points.

STEP 2:

List the boundary points of R where f has local maximums or minimums, and evaluate f (x, y) at these points.

STEP 3:

Look through the list for maximum and minimum values. These are the absolute extrema.

EXAMPLE 6:

Find the absolute maxima and minima of the function D (x, y) = x ² - xy + y ² + 1 on the closed triangular plate in the first quadrant bounded by the lines x = 0, y = 4, and y = x.

SOLUTION:

Here is a graph of the region that we are talking about in the problem.

So (0, 0) is a critical point.

First of all, the vertices of this triangle are also critical points because the ordinary derivative does not exist at these points.

So far the critical points are (0, 0), (0, 4), and (4, 4).

Now to determine if there are any critical points on the boundary.

SEGMENT OA

(0, 0) is a critical point.

SEGMENT AB

(2, 4) is a critical point.

SEGMENT OB

(0, 0) is a critical point.

Line segments OA and OB did not add any new critical points.

Here are the all the critical points and the function values at these points.

(0, 0)

f (0, 0) = 1

Absolute minimum

(0, 4)

f (0, 4) = 17

Absolute maximum

(2, 4)

f (2, 4) = 13

(4, 4)

f (4, 4) = 17

Absolute maximum

 EXAMPLE 7:

Find the absolute maxima and minima of the function T (x, y) = x ² + xy + y² - 6x on the rectangular plate 0 £ x £ 5, -3 £ y £ 3.

 SOLUTION:

Here is the graph of the region we are using.

 INTERIOR POINTS:

The point (4, -2) is a critical point.

Now to check the boundaries.

 SEGMENT AD

(0, 0) is a critical point.

SEGMENT AB

(-3/2, 3) is a critical point.

SEGMENT BC

(5, -5/2) is a critical point.

SEGMENT CD

(9/2, -3) is a critical point.

Here is the list of the critical points and their related function values.

(0, 3)

f (0, 3) = 9

(5, 3)

f (5, 3) = 19

Absolute maximum

(5, -3)

f (5, -3) = -11

(0, -3)

f (0, -3) = 9

(4, -2)

f (4, -2) = -12

Absolute minimum

(0, 0)

f (0, 0) = 0

(3/2, 3)

f (3/2. 3) = 6.75

(5, -5/2)

f (5, -5/2) = -11.25

(9/2, -3)

f (9/2, -3) = -11.25

 EXAMPLE 8:

Find the point on the graph z = x ² + y ² + 10 nearest the plane x + 2y - z = 0. 

SOLUTION:

We want the point on z = x ² + y ² + 10 where the tangent plane is parallel to x + 2y - z = 0.

Let w = z - x ² - y ² - 10, so Ñ w = -2x i -2y j + k is normal to z = x ² + y ² + 10 at (x, y). We need to find the values that will make Ñ w || n.

These values will give us the point farthest from the plane, so the point closest will be when x = 1/2 and y =1. The point is (1/2, 1, 45/4).

EXAMPLE 9:

Find the absolute maximum and minimum values of g (x, y) = xy on the quarter circle x ² + y ² = 4, x ³ 0, y ³ 0. (Use the parametric equations x = 2cos t, y = 2sin t.)

SOLUTION:

We will first need to find the interior critical points. To do this, we will have to find the derivative of g with respect to t. This will require us to use the chain rule for partial derivatives. If you need to review this method, please go to supplemental notes 16.

This is the only point that is in the quarter section.

The other two critical points occur when t = 0 and t = p /2. When t = 0, x = 2 and y = 0. When t = p /2, x = 0 and y = 2.