9.3

Maxima-Minima
OBJECTIVE
Find relative extrema of a function of two
variables.
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Inc. Publishing as Pearson AddisonWesley
DEFINITION:
A function f of two variables:
1. has a relative maximum at (a, b) if
f (x, y) ≤ f (a, b)
for all points in a rectangular region containing (a, b).
2. has a relative minimum at (a, b) if
f (x, y) ≥ f (a, b)
for all points in a rectangular region containing (a, b).
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Slide 6.3 - 2
THEOREM 1: The D-Test
To find the relative maximum and minimum values of f:
1. Find fx, fy, fxx, fyy, and fxy.
(find all 1st and 2nd order derivatives)
2. Solve the system of equations fx = 0, fy = 0. Let
(a, b) represent a solution. (set both 1st order derivatives = 0 and solve the
system to find (a, b) critical value)
3. Evaluate D, where D = fxx(a, b)·fyy(a, b) – [ fxy(a, b)]2.
(plug critical value (a, b) into D formula.)
continued
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Slide 6.3 - 3
THEOREM 1 (concluded):
4. Then: (4 possibilities)
a) f has a maximum at (a, b) if D > 0 and fxx(a, b) < 0.
b) f has a minimum at (a, b) if D > 0 and fxx(a, b) > 0.
c) f has neither a maximum nor a minimum at (a, b) if
D < 0. The function has a saddle point at (a, b).
d) This test is not applicable if D = 0.
In other words

If D is positive:
and fxx(a, b) is negative - There is a max at (a, b)
and fxx(a, b) is positive - There is a min at (a, b)

If D is negative:
If D = 0:
Saddle point at (a, b)
Test doesn’t apply.

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Slide 6.3 - 4
Example 1: Find the relative maximum or minimum
values of f (x, y)  x  xy  y  3x.
2
2
1. Find fx, fy, fxx, fyy, and fxy.
f x  2x  y  3
f y  x  2y
f xx  2
fyy  2
f xy  1
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Slide 6.3 - 5
Example 1 (continued):
2. Solve the system of equations fx = 0 and fy = 0.
2x  y  3  0
x  2y  0
x  2y
Using substitution,
2(2y)  y  3 
4y  y  3 
3y

y
0
0
3
 1.
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Slide 6.3 - 6
Example 1 (continued):
Then, substituting back,
x  2(1)
x 
2.
Thus, (2, –1) is the only critical point.
3. Find D.
D 


fxx (2,1)  fyy (2,1)  [ fxy (2,1)]2
2  2  [1]2
3
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Slide 6.3 - 7
Example 1 (concluded):
4. Since D = 3 and fxx(2, –1) = 2, since D > 0 and
fxx(2, –1) > 0, if follows from the D-Test that f has a
relative minimum at (2, –1). The minimum value is
f (2,1)  2 2  2(1)  (1)2  3 2


4  2 1 6
3
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Slide 6.3 - 8
Example 2: Find the relative maximum and
minimum values of
f (x, y)  xy  x 3  y2 .
1. Find fx, fy, fxx, fyy, and fxy.
fx  y  3x 2
fy  x  2y
fxx  6x
fyy  2
fxy  1
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Slide 6.3 - 9
Example 2 (continued):
2. Solve the system of equations fx = 0 and fy = 0.
y  3x 2  0
x  2y  0
y  3x 2
Using substitution,
x  2(3x 2 )  0
x  6x 2
 0
x(1  6x)  0
1
x  0 or x 
6
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Slide 6.3 - 10
Example 2 (continued):
Then, substitute back to find y for both values of x.
y  3(0)
y0
2
and
 1
y  3 
 6
and
1
y
12
2
Thus, (0,0) and (1/6, 1/12) are two critical points.
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Slide 6.3 - 11
Example 2 (continued):
3. Find D for both critical points. First, for (0,0)
D 
fxx (0, 0)  fyy (0, 0)  [ fxy (0, 0)]2

(6  0)  (2)  [1]2

1
Then, for (1/6, 1/12)
D 


1 1 
 1 1    1 1 
fxx  ,   fyy  ,    fxy  ,  
 6 12 
 6 12    6 12  
1

2
6


(2)

[1]


6
1
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2
Slide 6.3 - 12
Example 2 (concluded):
4. Since, for (0,0), D < 0, (0,0) is neither a maximum
nor a minimum value but a saddle point.
For (1/6, 1/12), D > 0 and fxx(1/6, 1/12) < 0.
Therefore, f has a relative maximum at (1/6, 1/12).
1 1 
f , 
 6 12 
3


1 1  1  1 
    
6 12  6   12 
1
1
1


72 216 144
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2

1
432
Slide 6.3 - 13
Example 3: A firm produces two kinds of golf balls,
one that sells for $3 and one priced at $2. The total
revenue, in thousands of dollars, from the sale of x
thousand balls at $3 each and y thousand at $2 each is
given by
R(x, y)  3x  2y.
The company determines that the total cost, in
thousands of dollars, of producing x thousand of the $3
ball and y thousand of the $2 ball is given by
C(x, y)  2x 2  2xy  y 2  9x  6y  7.
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Slide 6.3 - 14
Example 3 (continued):
How many balls of each type must be produced and
sold in order to maximize profit?
The total profit function P(x,y) is given by
P x, y   R(x, y)  C(x, y)
P x, y   3x  2y  (2x 2  2xy  y 2  9x  6y  7)
P x, y   2x 2  2xy  y 2  12x  4y  7
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Slide 6.3 - 15
Example 3 (continued):
1. Find Px, Py, Pxx, Pyy, and Pxy.
Px  4x  2y  12
Py  2x  2y  4
Pxx  4
Pyy  2
Pxy  2
2. Solve the system of equations Px = 0 and Py = 0.
4x  2y  12  0
2x  2y  4  0
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Slide 6.3 - 16
Example 3 (continued):
Adding these two equations, we get
Then,
2x  8  0.
2x  8
x
 4
Now, substitute back into Px = 0 or Py = 0 to find y.
2  4  2y  4  0
2y  4
 0
2y
 4
y

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2
Slide 6.3 - 17
Example 3 (continued):
Thus, (4, 2) is the only critical point.
3. Find D.
D  Pxx (4,2)  Pyy (4,2)  [Pxy (4,2)]2
D  4  2   2 
2
D4
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Slide 6.3 - 18
Example 3 (continued):
Thus, since D > 0 and Pxx (4,2) < 0, it follows that P
has a relative maximum at (4,2). So in order to
maximize profit, the company must produce and sell 4
thousand of the $3 golf ball and 2 thousand of the $2
golf ball. The maximum profit will be
P 4,2   2  4  2  4  2  2  12  4  4  2  7
2
2
P 4,2   13 or $13 thousand.
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Slide 6.3 - 19