Finding Maxima and Minima - University of British Columbia ...

Finding Maxima and Minima

When you were learning about derivatives about functions of one variable, you learned some techniques for finding the maximum and minimum values of functions of one variable. We'll now extend those techniques to functions of more than one variable. We'll concentrate on functions of two variables, though many of the techniques work more generally.

Local Maxima and Minima

One of the first things you did when you were developing the techniques used to find the maximum and minimum values of f (x) was you asked yourself

Suppose that the largest (or smallest) value of f (x) is f (a). What does that tell us about a?

After a little thought you answered

If the largest (or smallest) value of f (x) is f (a) and f is differentiable at a, then f (a) = 0.

Let's recall what that's true. Suppose that the largest value of f (x) is f (a). Then for all h > 0,

f (a + h) f (a)

=

f (a + h) - f (a) 0

=

f (a

+

h) h

-

f (a)

0

if h > 0

Taking the limit h 0 tells us that f (a) 0. Similarly, for all h < 0,

f (a + h) f (a)

=

f (a + h) - f (a) 0

=

f (a

+

h) h

-

f (a)

0

if h < 0

Taking the limit h 0 now tells us that f (a) 0. So we have both f (a) 0 and f (a) 0 which forces f (a) = 0. You also observed at the time that for this argument to work, you

only need f (x) f (a) for all x's close to a, not necessarily for all x's in the whole world. (In

the above inequalities, we only used f (a + h) with h small.) So you said

If f (a) is a local maximum or minimum for f (x) and f is differentiable at a, then f (a) = 0.

Exactly the same discussion applies to functions of more than one variable. Here are the corresponding definitions and statements.

Definition 1 (Local Max and Min).

The point (a, b) is a local maximum of the function f (x, y) if there is an r > 0 such that f (x, y) f (a, b) for all points (x, y) within a distance r of (a, b).

Similarly, (a, b) is a local minimum of the function f (x, y) if there is an r > 0 such that f (x, y) f (a, b) for all points (x, y) within a distance r of (a, b).

Local maximum and minimum values are also called extremal values.

c Joel Feldman. 2014. All rights reserved.

1

January 29, 2014

Definition 2 (Critical Point).

The point (a, b) is a critical point of the function f (x, y)

?

if

f x

(a,

b)

=

f y

(a,

b)

=

0

?

or

if

at

least

one

of

the

derivatives

f x

(a,

b),

f y

(a,

b)

does

not

exist.

Theorem 3.

If the function f (x, y) has local maximum or minimum at (a, b) and the partial

derivatives

f x

(a,

b),

f y

(a,

b)

exist,

then

f x

(a,

b)

=

f y

(a,

b)

=

0

Proof. It easy to see that this theorem follows from what we already know about functions of one variable. Suppose that f (x, y) has a local maximum or minimum at (a, b). Define the single variable functions

F (x) = f (x, b) G(y) = f (a, y)

Then a is a local maximum or minimum for the function F (x), so that F (a) = 0, and b is

a local maximum or minimum for the function G(y), so that G(b) = 0. Now we just have

to observe that F (x) is the rate of change of F (x) = f (x, y) y=b with respect to x when

y is held fixed at y = b, which is exactly

f x

(x,

b).

Similarly G(y) is the rate of change of

G(y)

=

f (x, y)

x=a

with

respect

to

y

when

x

is

held

fixed

at

x

=

a,

which

is

exactly

f y

(a,

y).

Thus

f x

(a,

b)

=

F

(a)

=

0

f y

(a,

b)

=

G(b)

=

0

Theorem 3 tells us that every local maximum or minimum is a critical point. Beware that it does not tell us that every critical point is either a local maximum or a local minimum. In fact, we shall see later, in Example 10, a critical point that is neither a local maximum nor a local minimum. None?the?less, Theorem 3 is very useful because often functions have only a small number of critical points. To find local maxima and minima of such functions, we only need to consider its critical points. We'll return later to the question of how to tell if a critical point is a local maximum, local minimum or neither. For now, we'll just practice finding critical points.

c Joel Feldman. 2014. All rights reserved.

2

January 29, 2014

Example 4 (f (x, y) = x2 - 2xy + 2y2 + 2x - 6y + 12) Find all critical points of f (x, y) = x2 - 2xy + 2y2 + 2x - 6y + 12. Solution. As a preliminary calculation, we find the two first order partial derivatives of f (x, y).

fx(x, y) = 2x - 2y + 2 fy(x, y) = -2x + 4y - 6

So the critical points are the solutions of the pair of equations 2x - 2y + 2 = 0, -2x + 4y - 6, or equivalently (dividing by two and moving the constants to the right hand side)

x - y = -1

(1a)

-x + 2y = 3

(1b)

One strategy for solving a system of two equations in two unknowns (x and y) like this is to

? First use one of the equations to solve for one of the unkowns in terms of the other unknown. For example (1a) tells us that y = x + 1. This expresses y in terms of x. We say that we have solved for y in terms of x.

? Then substitute the result, y = x + 1 in our case, into the other equation, (1b). In our case, this gives -x + 2(x + 1) = 3 x + 2 = 3 x = 1

? We have now found that x = 1, y = x + 1 = 2 is the only solution. So the only critical point is (1, 2).

An alternative strategy for solving a system of two equations in two unknowns like (1) is to

? add equations (1a) and (1b) together. This gives

(1a) + (1b) : (1 - 1)x + (-1 + 2)y = -1 + 3 y = 2

The point here is that adding equations (1a) and (1b) together eliminates the unknown x, leaving us with one equation in the unknown y, which is easily solved. For other systems of equations you might have multiply the equations by some numbers before adding them together.

? We now know that y = 2. Substituting it into (1a) gives us

x - 2 = -1 = x = 1

? Once again we have found that the only critical point is (1, 2).

Example 4

c Joel Feldman. 2014. All rights reserved.

3

January 29, 2014

Example 5 (f (x, y) = 2x3 - 6xy + y2 + 4y) Find all critical points of f (x, y) = 2x3 - 6xy + y2 + 4y. Solution. The first order partial derivatives are

fx = 6x2 - 6y fy = -6x + 2y + 4

So the critical points are the solutions of

6x2 - 6y = 0 - 6x + 2y + 4 = 0

We can rewrite the first equation as y = x2, which expresses y as a function of x. We can then substitute y = x2 into the second equation, giving

-6x + 2y + 4 = 0 -6x + 2x2 + 4 = 0 x2 - 3x + 2 = 0 (x - 1)(x - 2) = 0 x = 1 or 2

When x = 1, y = 12 = 1 and when x = 2, y = 22 = 4. So, there are two critical points: (1, 1), (2, 4).

Example 5

Example 6 (f (x, y) = xy(5x + y - 15)) Find all critical points of f (x, y) = xy(5x + y - 15). Solution. The first order partial derivatives of f (x, y) = xy(5x + y - 15) are

fx(x, y) = y(5x + y - 15) + xy(5) = y(5x + y - 15) + y(5x) = y(10x + y - 15) fy(x, y) = x(5x + y - 15) + xy(1) = x(5x + y - 15) + x(y) = x(5x + 2y - 15)

The critical points are the solutions of fx(x, y) = fy(x, y) = 0 or

y(10x + y - 15) = 0

and x(5x + 2y - 15) = 0

(2)

The first equation, y(10x+ y -15) = 0, is satisfied if either of the two factors y, (10x+ y -15) is zero. So the first equation is satisfied if either of the two equations

y=0

(3a)

10x + y = 15

(3b)

is satisfied. The second equation, x(5x + 2y - 15) = 0, is satisfied if either of the two factors x, (5x + 2y - 15) is zero. So the first equation is satisfied if either of the two equations

x=0

(4a)

5x + 2y = 15

(4b)

is satisfied. So both critical point equations (2) are satisfied if one of (3a), (3b) is satisfied and in addition one of (4a), (4b) is satisfied. There are four possibilities:

c Joel Feldman. 2014. All rights reserved.

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January 29, 2014

? (3a) and (4a) are satisfied if and only if x = y = 0

? (3a) and (4b) are satisfied if and only if y = 0, 5x + 2y = 15 y = 0, 3x = 15

? (3b) and (4a) are satisfied if and only if 10x + y = 15, x = 0 y = 15, x = 0

? (3b) and (4b) are satisfied if and only if 10x + y = 15, 5x + 2y = 15. We can use, for

example,

the

second

of

these

equations

to

solve

for

x

in

terms

of

y:

x

=

1 5

(15

-

2y).

When we substitute this into the first equation we get 2(15 - 2y) + y = 15, which we

can

solve

for

y.

This

gives

-3y

=

15 - 30

or

y

=

5

and

then

x

=

1 5

(15

-

2

?

5)

=

1.

In conclusion, the critical points are (0, 0), (3, 0), (0, 15) and (1, 5). A more compact way to write what we have just done is

fx(x, y) = 0

and

fy(x, y) = 0

y(10x + y - 15) = 0

and

x(5x + 2y - 15) = 0

y = 0 or 10x + y = 15

and

x = 0 or 5x + 2y = 15

x = y = 0 or y = 0, x = 3 or x = 0, y = 15 or x = 1, y = 5

Example 6

Example 7

In a certain community, there are two breweries in competition, so that sales of each negatively affect the profits of the other. If brewery A produces x litres of beer per month and brewery B produces y litres per month, then the profits of the two breweries are given by

P

=

2x

-

2x2 + y2 106

Q

=

2y

-

4y2 + x2 2 ? 106

respectively. Find the sum of the two profits if each brewery independently sets its own production level to maximize its own profit and assumes that its competitor does likewise. Find the sum of the two profits if the two breweries cooperate so as to maximize that sum.

Solution. If A adjusts x to maximize P (for y held fixed) and B adjusts y to maximize Q (for x held fixed) then x and y are determined by

Px

=

2-

4x 106

=

0

Qy

=

2-

8y 2?106

=

0

=

x

=

1 2

106

=

y

=

1 2

106

=

P

+

Q

=

2(x

+

y)

-

1 106

5 2

x2

+

3y2

=

= 106

1+1-

5 8

-

3 4

=

5 8

106

On

the

other

hand

if

(A, B)

adjust

(x, y)

to

maximize

P

+

Q

=

2(x

+

y)

-

1 106

5 2

x2

+

3y2

,

then x and y are determined by

c Joel Feldman. 2014. All rights reserved.

5

January 29, 2014

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