Math 20C (Shenk). Summer, 2008. Homework 3.
Section 14.2: 1, 19
Section 14.3: 9, 11, 19, 47, 49, 59
Section 14.4: 11, 17, 31, 35, 39
Section 14.5: 5, 23, 31, 41
Section 14.7: 5, 7, 9
1.
Give equations of (a) the plane that has slope 3 in the positive x-direction and slope −5 in the
positive y-direction, and whose z-intercept is 10 and (b) the plane that has slope −6 in the
positive x-direction and slope 10 in the positive y-direction and contains the point (5, 6, 7).
2.
Find the linear function z = L(x, y) with the level curves in Figure 1.
y
10
1
20
2
x
−20
−10
0
FIGURE 1
3.
4
Values of a function z = P (x, y) are given in the next table. Is it possible that P is linear? If so,
give a formula for such a function.
x=0
x = 10
x = 20
y=4
−12
−16
−20
y=0
0
−4
−8
What is the maximum directional derivative of z = x5 + y 3 at (1, 2)? Give the unit vector in the
direction of the maximum derivative.
Answer: [Maximum directional derivative] = 13 •
5 12
[Unit vector in the direction of the maximum derivative] = h 13
, 13 i
SOLUTION FOR INSTRUCTORS:
5.
z(x, y) = x5 + y 3 • ∇z(x, y) = h5x4 , 3y 2 i • ∇z(1,
2) = h5(1√4 ), 3(22 )i = h5, 12i •
√
2
[Maximum directional derivative] = |h5, 12i| = 5 + 122 = 169 = 13 •
h5, 12i
5 12
[Unit vector in the direction of the maximum derivative] =
= h 13
, 13 i
|h5, 12i|
Give unit vectors in the directions in which the directional derivative of f (x, y) = x + sin(5y) at
(2, 0) are zero.
h5, −1i
Answer: u = ± √
26
SOLUTION FOR INSTRUCTORS:
∂
(5y) = 5 cos(5y) •
∂y
fy (2, 0) = 5 cos[5(0)] = 5 • ∇f (2, 0) = h1, 5i • Unit vectors in the perpendicular directions:
f (x, y) = x + sin(5y) • fx (x, y) = 1 • fy (x, y) = cos(5y)
1
Math 20C, Summer, 2008. Homework 3 p. 2
u=±
6.
(8/17/08)
h5, −1i
h5, −1i
h5, −1i
=± √
= ±p
|h5, −1i|
26
52 + 12
What are the critical points of f (x, y) = x4 + 32x − y 3 ?
Answer: Critical point: (−2, 0)
SOLUTION FOR INSTRUCTORS:
7.
fx = 4x3 + 32 = 4(x3 + 8)
x3 + 8 = 0
f = x + 32x − y •
• Solve
•
2
y 2 = 0.
fy = −3y
The first equation gives x = −2 and the second gives y = 0. • Critical point: (−2, 0)
4
3
Find the critical points of f (x, y) = 3xy 2 + x3 − 3x.
Answer: Critical points: (0, 1), (0, −1), (1, 0), (−1, 0)
SOLUTION FOR INSTRUCTORS:
8.
y 2 + x2 − 1 = 0
fx = 3y 2 + 3x2 − 3 = 3(y 2 + x2 − 1)
•
• Solve
f = 3xy + x − 3x •
xy = 0.
fy = 6xy
The second equation holds if x = 0 or y = 0.
For x = 0, the first equation is y 2 − 1 = 0 and has the solutions y = ±1. •
For y = 0, the first equation is x2 − 1 = 0 and has the solutions x = ±1. •
Critical points: (0, 1), (0, −1), (1, 0), (−1, 0)
2
3
G(x, y) = 2x2 − x4 + 10y − y 2 has a global maximum. What is it and where does it occur?
Answer: [Global maximum] = 26 at (±1, 5)
SOLUTION FOR INSTRUCTORS:
9.
Since G is defined for all (x, y), its global
maximum is a local maximum and must
be at a critical
3
2
Gx = 4x − 4x = 4x(1 − x )
x(1 − x2 ) = 0
point. • G = 2x2 − x4 + 10y − y 2 •
• Solve
Gy = 10 − 2y = 2(5 − y)
5 − y = 0.
•
The first equation gives x = 0, 1, −1 and the second gives y = 5. • Critical points:
(0, 5), (1, 5), (−1, 5) • G(0, 5) = 2(0)2 − (0)4 + 10(5) − (5)2 = 25, G(1, 5) = 2(1)2 − (1)4 +
10(5) − (5)2 = 26,
G(−1, 5) = 2(−1)2 − (−1)4 + 10(5) − (5)2 = 26 • [Global maximum] = 26 at (±1, 5)
A three-dimensional rectangular frame is to be constructed with wire that costs $1.00 per meter
for the four horizontal edges parallel to the front, $0.75 per meter for the four horizontal edges
parallel to the sides, and $1.50 per meter for the four vertical edges. The volume of the box
enclosed by the frame is to be 24 cubic meters What dimensions minimize the cost?
Answer: The frame should be 3 meters wide, 4 meters deep and 2 meters high.
SOLUTION FOR INSTRUCTORS:
dollars
[4x meters]
[Width] = x, [Length] = y, [Height] = z • [Cost] = 1.00
meter
dollars
dollars
+ 0.75
[4y square feet] + 1.50
[4z square feet] = 4x + 3y + 6z dollars •
meter
meter
24
144
[Volume] = xyz • [Volume] = 24 =⇒ z =
• [Cost] = C(x, y) = 4x + 3y +
xy
xy
144
144
144
144
Cx = 4− 2 • Cy = 3− 2 + 2x • Cx = 0 ⇐⇒ 4x−
= 0 • Cy = 0 ⇐⇒ 3y −
=0
xy
xy
x y
xy
24
144
24
=
=2
• 4x = 3y • y = 43 x • 4 = 4 3 • x3 = 27 • x = 3 • y = 34 (3) = 4 • z =
xy
3(4)
3x
The frame should be 3 meters wide, 4 meters deep and 2 meters high.
(8/17/08)
10.
Math 20C, Summer, 2008. Homework 3, p. 3
Find the point on the plane z = 2x + 3y that is closest to the point (4, 2, 0). (Minimize the square
of the distance from (4, 2, 0) to a point on the plane.)
Answer: The closest point is (2, −1, 1)
SOLUTION FOR INSTRUCTORS:
11.
Minimize
f (x, y) = [Distance]2 = (x − 4)2 + (y − 2)2 + (2x + 3y)2 . •
5x + 6y − 4 = 0
fx = 2(x − 4) + 4(2x + 3y) = 10x + 12y − 8 = 2(5x + 6y − 4)
•
• Solve
3x + 5y − 1 = 0.
fy = 2(y − 2) + 6(2x + 3y) = 12x + 20y − 4 = 4(3x + 5y − 1)
Multiply the first equation by 5 and the second by −6 and add,. • (25−18)x+(30−30)y−20+6 =
0 • 7x = 14 • x = 2 • Use the first equation. • 5(2) + 6y − 4 = 0 • y = −1 •
z = 2x + 3y = 2(2) + 3(−1) = 1 • The closest point is (2, −1, 1).
Find the critical points of f = x3 − y 3 − 3xy and use the Second-Derivative Test to classify them.
The graph of the function is in Figure 2 and its level curves are in Figure 3.
FIGURE 2
WC: FIGURE 3
SOLUTION FOR INSTRUCTORS:
f = x3 − y 3 − 3xy • fx = 3x2 − 3y = 3(x2 − y), fy = −3y 2 − 3x = −3(y 2 + x) • Solve
x2 − y = 0
y 2 + x = 0.
The first equation gives y = x2 and then the second gives • x = −y 2 = −x4 . • x(x3 + 1) = 0 •
x = 0 and y = x2 = 0 or x = −1 and y = x2 = 1 •
Critical points: (0, 0) and (−1, 1) • fxx = 6x, fxy = −3, fyy = −6y • See the table.
Critical point
A = fxx
B = fxy
C = fyy
AC − B 2
Type
(0, 0)
0
−3
0
(0)(0) − (−3)2 = −9
Saddle point
(−1, 1)
−6
−3
−6
(−6)(−6) − (−3)2 = 27
Local maximum
Math 20C, Summer, 2008. Homework 3 p. 4
12.
(8/17/08)
Show that the Second-Derivative Test fails at the critical point of h(x, y) = 31 y 3 − x2 y. Its graph
is the “monkey saddle” in Figure 4. Its level curves are in Figure 5.
FIGURE 4
FIGURE 5
SOLUTION FOR INSTRUCTORS:
hx = −2xy, hy = y 2 − x2
equation requires x = 0 or y
first equation gives x = 0. •
At (0, 0): A = B = C = 0 •
• At critical points x2 − y 2 = 0 and −2xy = 0 • The second
= 0 • For x = 0, the first equation gives y = 0 • For y = 0, the
Critical point; (0, 0) • hxx = −2y, hxy = −2x, hyy = 2y •
AC − B 2 = 0 and the Second-Derivative Test fails.