4.7 Maximum and Minimum Values
Definition
A function of two variables has a local maximum at ( , ) if ( , )
≤ ( , ) when ( , ) is near ( , ). The number ( , ) is called a local maximum value. If ( , ) ≥ ( , ) when ( , ) is near ( , ), then has a local minimum at ( , ) and ( , ) is a local minimum value.
A point ( , ) is called a critical point (or stationary point) of if
, = 0 and , = 0.
Second Derivatives Test Suppose the second partial derivatives of are continuous on a disk with center ( , ), and suppose that ,
= 0 and , = 0 [that is, ( , ) is a critical point of ]. Let
= , = = −
(a) If > 0 and , > 0, then ( , ) is a local minimum.
(b) If > 0 and , < 0, then ( , ) is a local maximum.
(c) If < 0, then ( , ) is not a local maximum or minimum, the point ( , ) is called a saddle point.
NOTE If = 0, the test gives no information.
EXAMPLE 3 Find the local maximum and minimum values and saddle points of
, = + − 4 + 1
SOLUTION We first locate the critical points:
= 4 − 4 , = 4 − 4
Setting these partial derivatives equal to 0, we obtain the equations
= , =
To solve these equations we substitute = from the first equation into the second one. This gives
0 = − = ( − 1) = ( − 1)( + 1)
so there are three real roots: = 0,1, −1. The three critical points are (0,0), (1,1), (−1, −1).
Next we calculate the second partial derivatives and ( , ):
= 12 , = 12 , = −4
= , = = 144 − 16
Since 0,0 = −16 < 0, it follows from case (c) of the Second Derivatives Test that the origin is a saddle point; that is, has no local maximum or minimum at 0,0 .
Since 1,1 = 128 > 0, and (1,1) = 12, we see from case (a) of the test that (1,1) = −1 is a local minimum. Similarly, we have −1, −1 = 128
> 0, and (−1, −1) = 12, so (−1, −1) = −1 is also a local minimum.
Exercises: Find the local maximum and minimum values and saddle point(s) of the function.
(6) , = − 2 − 2 − −
SOLUTION The critical points:
= − 2 − 2
,
= − 2 − 2Setting these partial derivatives equal to 0, we obtain the equations 2 − = −2,
− 2 = 2
To solve these equations we multiply the first equation by 2 and subtracting the second we get = −2. The three critical point is (−2, −2).
The second partial derivatives and ( , ):
= −2, = −2, = 1
= , = = 4 − 1 = 3
Since −2, −2 = 3 > 0 , and −2, −2 = −2 , we see that
−2, −2 = 4 is a local maximum.
(7) , = − 1 − = − − + SOLUTION
The critical points:
= 1 − 2 +
,
= −1 − + 2Setting these partial derivatives equal to 0, we obtain 1 − 2 + = 0,
−1 + 2 − = 0
Added these two equations we get
= ⇒ = ± If = , then from the first equation we get
1 − 2 + = 0, ⇒ = 1 ⇒ = ±1
so there are two real roots: = 1, −1 . The critical points are (1,1) , (−1, −1).
If = − , then from the first equation we get 1 + 2 + = 0, which has no solution.
The second partial derivatives and ( , ):
= 1 − 2 + , = −1 − + 2
= −2 , = 2 , = 2 − 2
= , = = −4 − (2 − 2 )
Since 1,1 = −4 < 0, we see that the point 1,1 is a saddle point; that is, has no local maximum or minimum at 1,1 .
Also, −1, −1 = −4 < 0, we see that the point −1, −1 is a saddle point;
that is, has no local maximum or minimum at −1, −1 .
4.8
Lagrange Multipliers
Method of Lagrange Multipliers To find the maximum and minimum values of ( , , ) subject to the constraint ( , , ) = :
(a) Find all values of , , , and such that
, , = , ,
and
( , , ) =
(b) Evaluate at all the points ( , , ) that result from step (a). The largest of these values is the maximum value of ; the smallest is the minimum value of .
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If we write the vector equation , , = , , in terms of components, then the equations in step (a) become
= , = , =
EXAMPLE 3 Find the extreme values of the function , = + 2 on the circle + = 1.
SOLUTION We are asked for the extreme values of subject to the constraint g x, y = + = 1.
Using Lagrange multipliers, we solve the equations = and g x, y = 1, which can be written as
= , = , g x, y = 1
or as
2 = 2 (1) 4 = 2 (2)
+ = 1 (3)
From (1) we have = 0 or = 1. If = 0, then (3) gives = ±1. If
= 1, then = 0, from (2), so then (3) gives = ±1. Therefore has possible extreme values at the points (0,1), (−1,0), (0,1), and (0, −1). Evaluating at these four points, we find that
0,1 = 2, 0, −1 = 2, 1,0 = 1 , −1,0 = 1 Therefore the maximum value of on the circle + = 1 is
0, ±1 = 2 and the minimum value is ±1,0 = 1.
Exercises: Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint.
(4) , = 3 + , + = 10.
SOLUTION
Using Lagrange multipliers, we solve the equations = and g x, y = 10, which can be written as
= , = , g x, y = 1
or
3 = 2 (1) 1 = 2 (2)
+ = 10 (3)
Substitute from equations (1) , (2) by = , = in (3), we get + = 10 ⇒ 4 = 1 or = ± . So = ±3, = ±1.
Therefore has possible extreme values at the points (3,1), (3, −1), (−3,1), and (−3, −1). Evaluating at these four points, we find that
3,1 = 10, 3, −1 = 8
−3,1 = −8, −3, −1 = −10
Therefore the maximum value of on the circle + = 10 is 3,1 = 10 and the minimum value is −3, −1 = −10.
(7) , , = 2 + 2 + , + + = 9.
SOLUTION We solve the following equations = , = , = , g , , = 9 or 2 = 2 (1)
2 = 2 (2)
1 = 2 3
+ + = 9 (4)
Substitute from equations (1) , (2), (3) by = , = , =
in (4), we get + + = 9 ⇒ 4 = 1 or = ± .
So = ±2, = ±2, = ±1.
Therefore has possible extreme values at the eight points 2,2,1 , 2,2, −1 , 2, −2,1 , 2, −2, −1 , −2,2,1 , −2,2, −1
−2, −2,1 , and (−2, −2, −1). Evaluating at these eight points, we find that
2,2,1 = 9 , 2,2, −1 = 7, 2, −2,1 = 1, 2, −2, −1 = −1, −2,2,1 = 1, −2,2, −1 = −1
−2, −2,1 = −7, −2, −2, −1 = −9.
Therefore the maximum value of on the sphere + + = 9 is 2,2,1 = 9 and the minimum value is −2, −2, −1 = −9.
