LECTURE 8: MAXIMUM/MINIMUM PROBLEMS
MINGFENG ZHAO
January 21, 2015
Absolute maximum and minimum values
Let R be a bounded closed subset in R2 , to find the absolute maximum and minimum values of f on R:
Step 1: Find the maximum and minimum values of f on the boundary of R.
Step 2: Determine the values of f at all critical points in R.
Step 3: The greatest function value found in Step 1 and Step 2 is the absolute maximum value of f on R, and the
least function value found in Step 1 and Step 2 is the absolute minimum value of f on R.
Example 1. Find the absolute maximum and minimum values of f (x, y) = xy − 8x − y 2 + 12y + 160 over the triangular
region R := {(x, y) : 0 ≤ x ≤ 15, 0 ≤ y ≤ 15 − x}.
Step 1: Let O(0, 0), A(15, 0) and B(0, 15), then boundary of R consists of three parts, the line segment OA, the line
segment 0B and the line segment AB:
a. For all (x, y) on the line segment OA, we have y = 0 and 0 ≤ x ≤ 15, then
g1 (x) = f (x, 0) = x · 0 − 8x − 02 + 12 · 0 + 160 = 160 − 8x.
Now we should find the largest value and the smallest value of g1 (x) on 0 ≤ x ≤ 15. Since g10 (x) = −8 < 0,
then g1 (x) is a decreasing function for 0 ≤ x ≤ 15. So the largest value of f (x, y) on OA is equal to f (0, 0) =
g1 (0) = 160, and the smallest value of f (x, y) on OA is equal to f (15, 0) = g2 (15) = 160 − 8 · 14 = 40.
b. For all (x, y) on the line segment OB, we have x = 0 and 0 ≤ y ≤ 15, then
g2 (y) = f (0, y) = 0 · y − 8 · 0 − y 2 + 12y + 160 = −y 2 + 12y + 160.
Now we should find the largest value and the smallest value of g2 (y) on 0 ≤ y ≤ 15. Since g20 (x) = −2y + 12y,
then the critical point for g2 (y) is:
y = 6.
1
2
MINGFENG ZHAO
Notice that
g2 (0) = 160,
g2 (6) = 196,
and g2 (15) = 115
So the largest value of f (x, y) on OB is equal to f (0, 6) = g2 (6) = 196, and the smallest value of f (x, y) on
OB is equal to f (0, 15) = g2 (15) = 115.
c. For all point (x, y) on the line segment AB, we have y = 15 − x and 0 ≤ x ≤ 15, then
g3 (x) = f (x, 15 − x) = −2x2 + 25x + 115.
Now we should find the largest value and the smallest value of g3 (y) on 0 ≤ x ≤ 15. Since g30 (x) = −4x + 25,
then the critical point for g3 (x) is:
x=
25
.
4
Notice that
g3 (0) = 115,
g3 (25/4) = 193.125,
and g3 (15) = 40.
So the largest value of f (x, y) on AB is equal to f (25/4, 35/4) = g3 (25/4) = 193.125, and the smallest value
of f (x, y) on AB is equal to f (15, 0) = g3 (15) = 40.
Step 2: Now let’s find all critical points of f (x, y). In fact, we have
fx (x, y) = y − 8,
and fy (x, y) = x − 2y + 12.
So we should solve the following system:
y − 8 = 0,
and x − 2y + 12 = 0.
So x = 4 and y = 8, that is, f (x, y) has only one critical point (4, 8). Notice that
f (4, 8) = 192.
Step 3: By Step 1 and Step 2, we know that the largest value of f (x, y) on R is f (0, 6) = 196, the smallest value of
f (x, y) on R is f (15, 0) = 40.
Example 2. Find the absolute maximum and minimum values of f (x, y) = x2 + y 2 − 2x + 2y + 5 on the region
R := {(x, y) : x2 + y 2 ≤ 4}.
Step 1: Since R is a disc, then the boundary of R is the circle C = {(x, y) : x2 + y 2 = 4}. We can use the polar
coordinates:
x = 2 cos(θ),
and y = 2 sin(θ),
0 ≤ θ ≤ 2π.
LECTURE 8: MAXIMUM/MINIMUM PROBLEMS
3
Then
g(θ)
=
f (2 cos(θ), 2 sin(θ))
=
4 cos2 (θ) + 4 sin2 (θ) − 4 cos(θ) + 4 sin(θ) + 5
=
4 − 4 cos(θ) + 4 sin(θ) + 5
= −4 cos(θ) + 4 sin(θ) + 9,
Since cos2 (θ) + sin2 (θ) = 1
0 ≤ θ ≤ 2π.
So g 0 (θ) = 4 sin(θ) + 4 cos(θ). Let g 0 (θ) = 0, then tan(θ) = −1. Since 0 ≤ θ ≤ 2π, then θ = 34 π or θ =
g(0)
3
g
π
4
7
π
g
4
7π
4 .
Notice that
= −4 + 0 + 9 = 5
√
3
3
= −4 cos( π) + 4 sin( π) + 9 = 9 + 4 2
4
4
√
3
3
= −4 cos( π) + 4 sin( π) + 9 = 9 + 4 2.
4
4
√
√ √
√
√
3
7
Hence the largest value of f (x, y) on the boundary of R is 9+4 2 = g
π = f (− 2, 2) = g
π = f ( 2, − 2),
4
4
and the smallest value of f (x, y) on the boundary of R is 5 = g(0) = g(2, 0).
Step 2: Since fx (x, y) = 2x − 2 and fy (x, y) = 2y + 2, the we should solve the following system:
2x − 2 = 0,
and
2y + 2 = 0.
So x = 1 and y = −1, that is, we have only one critical point (1, −1). Notice that
f (1, −1) = 12 + 12 − 2 · 1 + 2 · (−1) + 5 = 3.
√
Step 3: In summary, we know that that absolute maximum value of f (x, y) on R is 9 + 4 2, and the absolute minimum
value of f (x, y) on R is 3.
Remark 1. In general, for the ellipse
x2
y2
+
= 1, one can use the polar coordinates:
a2
b2
x = a cos(θ),
and y = b sin(θ),
0 ≤ θ ≤ 2π.
Remark 2. In Friday’s class, we can use the Lagrange Multiplier to find the largest/smallest values for f (x, y) on R in
Example 2
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