LECTURE 6: MAXIMUM/MINIMUM PROBLEMS
MINGFENG ZHAO
January 16, 2015
Local maximum/minimum
Definition 1. A function f has a local maximum value at point (a, b) if f (x, y) ≤ f (a, b) for all (x, y) in some small
open disk centered at (a, b).
A function f has a local minimum value at point (a, b) if f (x, y) ≥ f (a, b) for all (x, y) in some small open disk
centered at (a, b).
Example 1. The function f (x, y) = x2 + y 2 − 4x + 2y + 5 has a local minimum at (2, −1).
In fact, we know that
f (x, y)
=
x2 − 4x + y 2 + 2y + 5
=
x2 − 4x + 4 − 4 + y 2 + 2y + 1 − 1 + 5
=
(x − 2)2 − 4 + (y + 1)2 − 1 + 5
=
(x − 2)2 + (y + 1)2 .
Completing the square
Now it’s easy to see that f (x, y) has a local minimum at (2, −1), and f (2, −1) = 0.
Remark 1. For completing the square:
b
2
ax + bx + c = a x + x + c
a
"
2 2 #
b
b
b
2
−
+c
= a x +2· x+
2a
2a
2a
"
#
2
b
b2
= a x+
− 2 +c
2a
4a
2
b
b2
= a x+
−a· 2 +c
2a
4a
2
1
2
MINGFENG ZHAO
2
b
b2
= a x+
−
+c
2a
4a
2
4ac − b2
b
+
= a x+
.
2a
4a
Critical points
Definition 2. A point (a, b) is called a critical point of f if fx (a, b) = fy (a, b) = 0.
Theorem 1. If f has a local maximum or local minimum value at point (a, b), then (a, b) is a critical point of f , that
is, fx (a, b) = fy (a, b) = 0.
Remark 2. But a critical point of F maybe neither a local maximum nor local minimum value at this critical point.
Example 2. In Example 1, we know that f (x, y) = x2 + y 2 − 4x + 2y + 5 has a local minimum at (2, −1), by Theorem
1, we have fx (2, −1) = fy (2, −1) = 0.
In fact, for partial derivatives of f , we have
fx (x, y) = 2x − 4,
and fy (x, y) = 2y + 2.
Then we get
fx (2, −1) = 2 · 2 − 4 = 0,
and fy (2, −1) = 2 · (−1) + 2 = 0.
Example 3. Find all critical points of f (x, y) = xy(x − 2)(y + 3).
For partial derivatives of f , we have
fx (x, y)
fy (x, y)
= y(x − 2)(y + 3) + xy(y + 3)
=
(xy − 2y)(y + 3) + xy(y + 3)
=
(xy − 2y + xy)(y + 3)
=
(2xy − 2y)(y + 3)
=
2y(x − 1)(y + 3)
= x(x − 2)(y + 3) + xy(x − 2)
=
(xy + 3x)(x − 2) + xy(x − 2)
By the product rule
By the product rule
LECTURE 6: MAXIMUM/MINIMUM PROBLEMS
=
(xy + 3x + xy)(x − 2)
=
(2xy + 3x)(x − 2)
3
= x(2y + 3)(x − 2).
To find critical points of f , we should solve the following system:
2y(x − 1)(y + 3) = 0
(1)
fx (x, y)
=
(2)
fy (x, y)
= x(2y + 3)(x − 2) = 0.
Let’s first solve the first equation (1), since 2y(x − 1)(y + 3) = 0, then 2y = 0 or x − 1 = 0 or y + 3 = 0, which gives
us:
y = 0,
or x = 1,
or y = −3.
When y = 0, plug y = 0 into the second equation (2), we get x(2 · 0 + 3)(x − 2) = 0, that is, 3x(x − 2) = 0, then
3x = 0 or x − 2 = 0, that is, x = 0 or x = 2. In this case, we get two critical points: (0, 0) and (2, 0).
When x = 1, plug x = 1 into the second equation (2), we get 1(2y + 3)(1 − 2) = 0, that is, −(2y + 3) = 0, then
2y + 3 = 0, that is, y = − 23 . In this case, we get one critical point: (1, − 23 ).
When y = −3, plug y = −3 into the second equation (2), we get x(2 · (−3) + 3)(x − 2) = 0, that is, −3x(x − 2) = 0,
then −3x = 0 or x − 2 = 0, that is, x = 0 or x = 2. In this case, we get two critical points: (0, −3) and (2, −3).
In summary, we get five critical points of f :
(0, 0),
(2, 0),
3
(1, − ),
2
(0, −3),
and
(2, −3).
Second derivative test for classifying critical points
Definition 3. A point (a, b) is call a saddle point of f if (a, b) is a critical point of f , and in every small open disk
centered at (a, b), there are points (x, y) for which f (x, y) > f (a, b) and points for which f (x, y) < f (a, b).
Definition 4. The Hessian matrix of f is defined as:

fxx
fxy
fyx
fyy


.
The discriminant of f , D(x, y), is defined as:
D(x, y) = fxx (x, y)fyy (x, y) − (fxy (x, y))2 .
4
MINGFENG ZHAO
Example 4. Let f (x, y) = x2 y 2 , compute the Hessian matrix and discriminant of f .
In fact, we have
fx = 2xy 2 ,
and fy = 2x2 y.
Then
fxx = 2y 2 ,
fxy = fyx = 4xy,
and fyy 2x2 .
So the Hessian matrix of f is:

2y 2
4xy
4xy
2x2


.
So the discriminant of f is:
D(x, y) = fxx fyy − (fxy )2 = 2y 2 · 2x2 − (4xy)2 = 4x2 y 2 − 16x2 y 2 = −12x2 y 2 .
Remark 3. The critical points for f (x, y) = x2 y 2 are (0, y) and (x, 0) for all x and all y, and D(x, 0) = D(0, y) = 0.
Theorem 2 (Second Derivative Test). Let the point (a, b) be a critical point of f , that is, fx (a, b) = fy (a, b) = 0, then
a. If D(a, b) > 0 and fxx (a, b) < 0, then f has a local maximum value at (a, b).
b. If D(a, b) > 0 and fxx (a, b) > 0, then f has a local minimum value at (a, b).
c. If D(a, b) < 0, then f has a saddle point at (a, b).
d. If D(a, b) = 0, then the test is inconclusive.
Example 5. For the function f (x, y) = x2 + y 2 − 4x + 2y + 5 in Example 1, by Example 2, we know that (2, −1) is
the only critical point for f (x, y). By the computations in Example 2, we know that
fxx (2, −1) = 2,
fxy (2, −1) = 0,
and fyy (2, −1) = 2.
So D(2, −1) = 2 · 2 − 02 = 4 > 0. Since fxx (2, −1) = 2 > 0, by the Second Derivative Test, we know that (2, −1) is a
local minimum for f (x, y).
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca