Section 12.7: Maximum and Minimum Values
Definition:
• A function f (x, y) has a local maximum at (a, b) if f (x, y) ≤ f (a, b) for all (x, y) in
some disk with center (a, b). The number f (a, b) is called a local maximum value.
Similarly, f (x, y) has a local minimum at (a, b) if f (x, y) ≥ f (a, b) for all (x, y) in
some disk with center (a, b). The number f (a, b) is called a local minimum value.
A local maximum or minimum is called a local extremum.
• A function f (x, y) has an absolute maximum at (a, b) if f (x, y) ≤ f (a, b) for all
(x, y) in the domain of f . The number f (a, b) is called an absolute maximum
value. Similarly, f (x, y) has an absolute minimum at (a, b) if f (x, y) ≥ f (a, b) for
all (x, y) in the domain of f . The number f (a, b) is called an absolute minimum
value. An absolute maximum or minimum is called an absolute extremum.
Theorem: (Partial Derivative Criteria)
If f has a local extremum at (a, b) and the first-order partial derivatives of f exist at (a, b),
then fx (a, b) = fy (a, b) = 0.
Definition: A point (a, b) such that fx (a, b) = fy (a, b) = 0, or one of these partial derivatives
does not exist, is called a critical point of f .
Example: Find all critical points of f (x, y) = x2 + y 2 + 4x − 6y + 1.
The first-order partial derivatives are
fx (x, y) = 2x + 4
and
fy (x, y) = 2y − 6.
Setting these partial derivatives equal to zero, we have x = −2 and y = 3. Thus, (−2, 3) is
the only critical point.
Example: Find all critical points of f (x, y) = 2x2 + y 2 + 2xy + 2x + 2y.
The first-order partial derivatives are
fx (x, y) = 4x + 2y + 2
and
fy (x, y) = 2y + 2x + 2.
Setting these partial derivatives equal to zero, we have the system
4x + 2y + 2 = 0,
2y + 2x + 2 = 0.
It follows that x = 0 and y = −1. Thus, (0, −1) is the only critical point.
Example: Find all critical points of f (x, y) = 2x3 + xy 2 + 5x2 + y 2 .
The first-order partial derivatives are
fx (x, y) = 6x2 + y 2 + 10x
and
fy (x, y) = 2xy + 2y.
Setting these partial derivatives equal to zero, we have the system
6x2 + y 2 + 10x = 0,
2y(x + 1) = 0.
It follows from the second equation that either x = −1 or y = 0. If x = −1, then the first
equation becomes y 2 − 4 = 0 and so y = ±2. Thus, (−1, −2) and (−1, 2) are critical points.
If y = 0, then the first equation becomes 6x2 + 10x = 2x(3x + 5) = 0 and so x = 0 or
x = −5/3. Thus, (0, 0) and (0, −5/3) are critical points.
Definition: A critical point (a, b) of f is called a saddle point if every disk centered at (a, b)
contains points (x, y) such that f (a, b) ≤ f (x, y) and points such that f (a, b) ≥ f (x, y).
Theorem: (Second Partials Test)
Suppose f is a function with continuous second-order derivatives and suppose (a, b) is a
critical points of f . Let
D = D(a, b) = fxx (a, b)fyy (a, b) − [fxy (a, b)]2 .
1. If D > 0 and fxx (a, b) > 0, then f (a, b) is a local minimum.
2. If D > 0 and fyy (a, b) < 0, then f (a, b) is a local maximum.
3. If D < 0, then (a, b) is a saddle point.
Note: The function D is called the discriminant. If D = 0, the Second Partials Test gives
no information.
Example: Find the local extrema and saddle points of f (x, y) = 3x2 + 8y 3 + 12x − 12y 2 + 7.
The first-order partial derivatives are
fx (x, y) = 6x + 12
and
fy (x, y) = 24y 2 − 24y.
Setting these partial derivatives equal to zero, we have the system
6(x + 2) = 0,
24y(y − 1) = 0.
Thus, the critical points are (−2, 0) and (−2, 1).
The second-order partial derivatives are
fxx (x, y) = 6
fxy = 0
fyy = 48y − 24.
Thus, the discriminant is
D(x, y) = (6)(48y − 24) − 02 = 288y − 144.
Since D(−2, 0) = −144 < 0, (−2, 0) is a saddle point. Since D(−2, 1) = 144 > 0 and
fxx (−2, 1) = 6 > 0, f (−2, 1) = −3 is a local minimum.
Example: Find the local extrema and saddle points of f (x, y) = x3 − 3xy + y 3 .
The first-order partial derivatives are
fx (x, y) = 3x2 − 3y
and
fy (x, y) = 3y 2 − 3x.
Setting these partial derivatives equal to zero, we have the system
3(x2 − y) = 0,
3(y 2 − x) = 0.
It follows that y = x2 and
y 2 − x = x4 − x = x(x2 − 1) = x(x + 1)(x − 1) = 0.
This equation has three roots: x = 0, −1, 1. Thus, there are three critical points (0, 0), (1, 1),
and (−1, −1).
The second-order partial derivatives are
fxx (x, y) = 6x
fxy = −3
fyy = 6y.
Thus, the discriminant is
D(x, y) = (6x)(6y) − (−3)2 = 36xy − 9.
Since D(0, 0) = −9 < 0, (0, 0) is a saddle point. Since D(1, 1) = 27 > 0 and fxx (1, 1) = 6 > 0,
f (1, 1) = −1 is a local minimum. Since D(−1, −1) = 27 > 0 and fxx (−1, −1) = −6 < 0,
f (−1, −1) = −5 is a local maximum.
Example: Find the points on the surface z 2 = xy + 1 that are closest to the origin.
The distance from a point (x, y, z) on the surface to the origin is given by
p
p
d = x2 + y 2 + z 2 = x2 + y 2 + xy + 1.
The first-order partial derivatives are
2x + y
dx = p
2 x2 + y 2 + xy + 1
and
2y + x
dy = p
.
2 x2 + y 2 + xy + 1
Setting these partial derivatives equal to zero, we have y = −2x and
2y + x = 2(−2x) + x = −3x = 0.
It follows that x = y = 0 and so the points of interest are (0, 0, 1) and (0, 0, −1).
Example: A cardboard box without a lid must have a volume of 23, 328 cm3 . Find the
dimensions that minimize the amount of cardboard used.
Let x, y, and z denote the dimensions of the box (in cm).
𝑦
𝑧
𝑥
The volume of the box is V = xyz = 23, 328 and so z = 23, 328/xy. Thus, the surface area
of the box is given by
S = xy + 2yz + 2xz = xy +
46, 656 46, 656
+
.
x
y
The first-order partial derivatives are
Sx = y −
46, 656
x2
and
Sy = x −
46, 656
.
y2
Setting these partial derivatives equal to zero, we have y = 46, 656/x2 and so
46, 656
x3
x4
x−
=x 1−
= 0.
=x−
y2
46, 656
46, 656
√
Clearly x 6= 0 and so x = 3 46, 656 = 36. Thus, y = 36 and z = 18. The dimensions of the
box with smallest surface area are 36 × 36 × 18.
Definition: A closed set in R2 is a set that contains all of its boundary points. A bounded
set in R2 is a set that is contained in some disk.
Theorem: (Extreme Value Theorem)
If f is continuous on a closed, bounded set D in R2 , then f attains an absolute maximum
and minimum value on D.
To find the absolute extrema of a continuous function f on a closed, bounded set D:
1. Find all extreme values in D.
2. Find all extreme values on the boundary of D.
3. The largest of these values is the absolute maximum and the smallest is the absolute
minimum.
Example: Find the absolute extrema of f (x, y) = 2+2x+2y −x2 −y 2 on the closed triangular
region T with vertices (0, 0), (0, 4), and (4, 0).
Since f is a polynomial, it is continuous on the closed, bounded set T . The first-order partial
derivatives are
fx (x, y) = 2 − 2x
and
fy (x, y) = 2 − 2y.
Thus, the only critical point in the interior of T is (1, 1) and f (1, 1) = 4.
The boundary of T consists of three line segments L1 , L2 , L3 .
0,4
𝐿3
𝑦 =4−𝑥
𝐿2
0,0
𝐿1
4,0
On L1 , we have y = 0 and
f (x, 0) = 2 + 2x − x2 ,
0 ≤ x ≤ 2.
Therefore f 0 (x) = 2 − 2x = 0 and so x = 1. Thus, (1, 0) is a critical point and f (0, 0) = 2,
f (1, 0) = 3, f (4, 0) = −6.
On L2 , we have x = 0 and
f (0, y) = 2 + 2y − y 2 ,
0 ≤ y ≤ 2.
Therefore f 0 (y) = 2 − 2y = 0 and so y = 1. Thus, (0, 1) is a critical point and f (0, 0) = 2,
f (0, 1) = 3, f (0, 4) = −6.
On L3 , we have y = 4 − x and
f (x, 4 − x) = 2 + 2x + 2(4 − x) − x2 − (4 − x)2 = 8x − 2x2 − 6.
Therefore, f 0 (x) = 8 − 4x = 0 and so x = 2. Thus, (2, 2) is a critical point and f (2, 2) = 2.
The absolute maximum value is f (1, 1) = 4 and the absolute minimum value is f (4, 0) =
f (0, 4) = −6.