LECTURE 7: MAXIMUM/MINIMUM PROBLEMS
MINGFENG ZHAO
January 19, 2015
Second derivative test for classifying critical points
Theorem 1 (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.
Where D(a, b) := fxx (a, b)fyy (a, b) − (fxy (a, b))2 is the discriminant of f .
Example 1. Use the Second Derivative Test to classify the critical points of f (x, y) = xy(x − 2)(y + 3).
By Example 3 in Lecture 6, we know that all critical points of f are:
(0, 0),
(2, 0),
3
(1, − ),
2
(0, −3),
and
(2, −3).
Also, we have
fx = 2y(x − 1)(y + 3),
and fy = x(2y + 3)(x − 2).
In order to use the Second Derivative Test, we need to compute the second derivatives. So
fxx
=
2y(y + 3)
fxy
=
fyx = 2(x − 1)(y + 3) + 2y(x − 1)
fyy
=
2x(x − 2).
Now let’s compute D(x, y) and fxx (x, y) at those critical points:
1
By the product rule
2
MINGFENG ZHAO
(x, y)
D(x, y) fxx
Conclusion
(0, 0)
−36
0
Saddle point
(2, 0)
−36
0
Saddle point
(1, − 32 )
9
− 92
Local maximum
(0, −3)
−36
0
Saddle point
(2, −3)
−36
0
Saddle point
Absolute maximum and minimum values
Definition 1. Let f (x, y) be defined on a subset R of R2 containing point (a, b), then
I. we say that f (a, b) is an absolute maximum value of f on R if f (a, b) ≥ f (x, y) for all (x, y) in R.
II. we say that f (a, b) is an absolute minimum value of f on R if f (a, b) ≤ f (x, y) for all (x, y) in R.
Remark 1. Any function on a bounded closed subset R will attain its absolute maximum value and the absolute
minimum values on R
Remark 2. 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 2. A shipping company handles rectangular boxes provided the sum of the length, width, and height of the
box does not exceed 96 inches. Find the dimensions of the box that meets this condition and has the largest volume.
Let
x
=
the length of the box
y
=
the width of the box
z
=
the height of the box
V
=
the volume of the box.
LECTURE 7: MAXIMUM/MINIMUM PROBLEMS
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By the assumption, we know that 0 ≤ x, y, z ≤ 96, V = xyz and x + y + z ≤ 96. In order to have the largest volume,
we should have x + y + z = 96, then z = 96 − x − y, which implies that
V = xy(96 − x − y).
By the definition of V , it’s easy to see that V (x, y, z) = 0 if one of x, y and z is 0. Therefore, V must attain its
absolute maximum interior which will be a critical point. Let’s find critical points of V :
Vx
Vy
=
y(96 − x − y) + xy · (−1)
=
y(96 − x − y) − xy
=
y(96 − x − y − x)
=
y(96 − 2x − y)
=
x(96 − x − y) + xy · (−1)
=
x(96 − x − y) − xy
=
x(96 − x − y − y)
=
x(96 − x − 2y).
So we should solve the following system:
(1)
y(96 − 2x − y)
=
0
(2)
x(96 − x − 2y)
=
0
First, let’s solve the first equation (1), we get either y = 0 or 96 − 2x − y = 0, that is, we have two cases: y = 0 or
y = 96 − 2x.
When y = 0, plug y = 0 into (2), we get x(96 − x − 0) = 0, that is, x(96 − x) = 0. Then x = 0 or x = 96. So we have
two critical points (0, 0) and (96, 0).
When y = 96 − 2x, plug y = 96 − 2x into (2), we get x[96 − 2(96 − 2x)] = 0, that is, x[96 − x − 2 · 96 + 4x] = 0. Then
x[3x − 96] = 0. Hence x = 0 or x = 32. Since y = 96 − 2x, then y = 96 or x = 96 − 2 · 32 = 32. So we have two critical
points (0, 96) and (32, 32).
In summary, we have four critical points:
(0, 0),
(96, 0),
(0, 96),
and
(32, 32).
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MINGFENG ZHAO
But we only need the critical points which are interior, so V will attain its absolute maximum value at (32, 32). In
this case, x = y = 32, z = 96 − x − y = 96 − 32 − 32 = 32, and V = xyz = 323 = 32, 768.
Remark 3. Also we can use the Second Derivative Test to check (32, 32) is a local maximum point for V . In fact,
Vxx = −2y,
Vxy = 96 − 2x − y − y = 96 − 2x − 2y,
and Vyy = −2x.
Then we have
Vxx (32, 32)
D(32, 32)
= −2 · 32 = −64 < 0
=
(−2 · 32) · (−2 · 32) − (96 − 2 · 32 − 2 · 32)2
=
3072 > 0.
So (32, 32) is a local maximum for V .
Example 3. 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: The 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.
Notice that
g2 (0) = 160,
g2 (6) = 196,
and g2 (15) = 115
LECTURE 7: MAXIMUM/MINIMUM PROBLEMS
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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
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.
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