Math 16B: Homework 2
Due: July 9
1. (a) f (x, y) = exy subject to x3 + y 3 = 16:
We define F (x, y, λ) = exy + λ(x3 + y 3 − 16). Then, at an extremum, we must
have:
∂F
= yexy + 3λy 2 = 0
∂x
∂F
= xexy + 3λx2 = 0
∂y
∂F
= x3 + y 3 − 16 = 0
∂λ
= −e3y . Plug this into the second
From the first equation, we have: λ = −ye
3y 2
equation to get:
( xy ) (
)
−e
x2 xy
xy
2
0 = xe + 3x
= x−
e
3y
y
)
(
2
2
Since exy ̸= 0, it follows that x − xy = 0 ⇒ y = xx = x. The third equation
then shows that
x3 + x3 = 16 ⇒ x3 = 8 ⇒ x = 2
xy
xy
It follows that y = 2. Thus, the extremum occurs at (2, 2).
(b) f (x, y, z) = xyz subject to x2 + 2y 2 + 3z 2 = 6:
We define F (x, y, z, λ) = xyz + λ(x2 + 2y 2 + 3z 2 − 6). Then, at an extremum, we
must have:
∂F
= yz + 2λx = 0
∂x
∂F
= xz + 4λy = 0
∂y
∂F
= xy + 6λz = 0
∂z
∂F
= x2 + 2y 2 + 3z 2 − 6 = 0
∂λ
. Plug this into the second to get
From the first equation, we get λ = −yz
2x
(
(
)
)
−yz
2y 2
2y 2 z
0 = xz + 4y
=z x−
= xz −
2x
x
x
1
(
This shows that either z = 0 or x −
2y 2
x
)
= 0. In the former case, the third and
√
√
fourth equations imply that either x = 0, 2y 2 = 6 ⇒
y
=
3
or
y
=
0,
x
=
6.
√
√
Thus, if z = 0, the objective function becomes f (0, 3, 0) = f ( 6, 0, 0) = 0.
(
)
2
In the case that x − 2yx = 0, we get by rearrangement x2 = 2y 2 . Plugging the
expression for λ in the third equation gives:
(
)
(
)
(
)
−yz
3yz 2
3z 2
0 = xy + 6z
= xy −
=y x−
2x
x
x
(
)
2
Hence, either y = 0 or x − 3zx =. By x2 = 2y 2 then, x = 0; the fourth equation
√
√
then shows √
that 3z 2 = 6 ⇒ z = ± 2. Thus, we may have extrema at (0, 0, 2)
and (0, 0, − 2); the value of f at both of these is 0.
)
(
2
In the case that x − 3zx =, we get by rearrangement 3z 2 = x2 . Finally, using
the facts that x2 = 2y 2 = 3z 2 in the fourth equation, we have
√
0 = x2 + x2 + x2 − 6 ⇒ x = ± 2
√
This corresponds to y = ±1 and z = ± 23 . We therefore get 8 extrema:
√ ) (√
√ ) (√
√ ) (√
√ ) ( √
√ )
(√
2, 1, 23 ,
2, −1, 23 ,
2, 1, − 23 ,
2, −1, − 23 , − 2, 1, 23 ,
√ ) ( √
√ )
√ )
( √
( √
2
2
− 2, −1, 3 , − 2, 1, − 3 and − 2, −1, − 23 . The resultant values
of f are ± √23 with the sign dependent on how many of x, y and z are negative.
2. This problem requires us to find x, y and z such that compared to all other points on
the planet, they represent the point with the lowest temperature. Thus, we need to
minimize T subject to x2 + y 2 + z 2 = 62 . Define the function
F (x, y, z, λ) = (6y − x2 + yz + 40) + λ(x2 + y 2 + z 2 − 36)
At an extremum, we must have:
∂F
= −2x + 2λx = 0
∂x
∂F
= 6 + z + 2λy = 0
∂y
∂F
= y + 2λz = 0
∂z
∂F
= x2 + y 2 + z 2 − 36 = 0
∂λ
From the first equation, we get 2x(−1 + λ) = 0 ⇒ either x = 0 or λ = 1. In the former
. Plug this into the third equation to
case, the second equation implies that λ = −6−z
2y
get
(
)
(
)
−6 − z
6+z
0 = y + 2z
=y−z
2y
y
2
Rearranging gives y 2 = z 2 − 6z. Finally, the last equation gives:
36 = 0 + z 2 − 6z + z 2 ⇒ z 2 − 3z − 18 = 0 ⇒ (z − 6)(z + 3) = 0
√
√
Thus,√z = −3 or z = 6. In the former case, y 2 = 27 ⇒ y = ± 27 so (0, 27, −3) and
(0, − 27, −3) are possible extrema. In the case z = 6, we get y 2 = 36−36 = 0 ⇒ y = 0
so (0, 0, 6) is a possible extremum.
In the case λ = 1, we get from the second and third equations
2y + z = −6
y + 2z = 0
Solve these simultaneously to get y = −4 and z = 2. Finally, using the constraint
equation gives
x2 + 16 + 4 = 36 ⇒ x2 = 16 ⇒ x = ±4
Thus, (4, −4, 2) and (−4, −4, 2) are two more possible extrema.
To find the coldest point, we find the temperature T (x, y, z) = 6y − x2 + yz + 40 at all
of the extrema we have found and pick the one(s) with the lowest value. We have:
√
√
T (0, 27, −3) = 40 + 3 27 ≈ 55.59
√
√
T (0, − 27, −3) = 40 − 3 27 ≈ 24.41
T (0, 0, 6) = 40
T (4, −4, 2) = −8
T (−4, −4, 2) = −8
Thus, the lowest temperature on the surface of the planet is shared by two points:
(4, −4, 2) and (−4, −4, 2).
3. (a) The region of integration is shown in Figure 1.
We have
]x 2
∫ 2 ∫ x2
∫ 2[
xy 4
2
3
2
dx
x + xy dy dx =
x y+
4 x
1
x
1
] [
]
∫ 2[
x(x2 )4
x(x)4
2
2
2
=
(x )(x ) +
− (x )(x) +
dx
4
4
1
∫ 2
x9
x5
=
x4 +
− x3 −
dx
4
4
1
[ 5
]2
x10
x4
x6
x
+
−
−
=
5
4(10)
4
4(6)
[ 5
] [1
]
10
4
6
2
2
2
1
1
1
2
1
+
−
−
+
− −
=
−
5
40
4
24
5 40 4 24
[
] [
]
32 128
8
1
1
1
1
=
+
−4−
+
− −
−
5
5
3
5 40 4 24
127
= 25.4
=
5
3
5
4.5
4
3.5
3
2
y
y=x
2.5
2
1.5
y=x
1
0.5
0.5
1
1.5
x
2
2.5
Figure 1: Region defined in Problem 3(a)
2.5
y = ln(8)
y
2
1.5
x = ln(y)
1
0.5
−0.5
0
0.5
x
Figure 2: Region defined in Problem 3(b)
4
1
(b) The region of integration is shown in Figure 2.
∫
ln 8
∫
∫
ln y
x+y
e
1
ln 8
dx dy =
0
∫
1
[ x+y ]ln y
e
dy
0
ln 8
eln y+y − ey dy
=
∫
1
ln 8
eln y ey − ey dy
=
∫
1
ln 8
yey − ey dy
1
∫ ln 8
8
y ln 8
= [ye ]1 −
(1)ey dy − [ey ]ln
1
=
1
8
y ln 8
= [yey ]ln
1 − 2 [e ]1
8
= [(y − 2)ey ]ln
1
[
]
= [(ln 8 − 2)(8)] − (−1)e1
= 8 ln 8 − 16 + e
(c) The region of integration is shown in Figure 3.
2.5
2
y = x1/2
y
1.5
1
0.5
0
y=0
−0.5
0.5
1
1.5
2
2.5
x
3
3.5
4
Figure 3: Region defined in Problem 3(c)
5
4.5
∫
1
4
∫
√
x
0
3
e
2
√y
x
[ ( √y )]√x
3
e x
√
dx
2 1/ x
1
0
]√x
∫ 4[ √
3 x √yx
e
dx
2
1
0
] [ √
]
∫ 4[ √
3 x 0
3 x 1
e −
e dx
2
2
1
∫ 4 √
3 x
(e − 1) dx
2
1
[ 3/2 ]4
3
x
(e − 1)
2
3/2 1
[ 3/2
]
(e − 1) 4 − 13/2
7(e − 1)
∫
dy dx =
=
=
=
=
=
=
4
3
2
y = e−x
y
1
0
−1
y = −e−x
−2
−3
−2
−1
0
1
2
3
4
5
x
Figure 4: Region defined in Problem 4
4. The region is shown in Figure 4. Observe that, since y ≥ −e−x , whenever (x, y) belongs
to the region of integration, z = y + ex + e−x ≥ ex > 0. Thus, the surface never dips
below the x-y plane so the required volume can be obtained by simply integrating the
6
surface over the region. We therefore get the volume
∫ 4 ∫ y=e−x
V =
y + ex + e−x dy dx
−1
∫
4
[
=
−1
∫ 4
[
=
−1
4
∫
=
y=−e−x
y2
+ y(ex + e−x )
2
]y=e−x
dx
y=−e−x
] [ −2x
]
e−2x
e
−x x
−x
−x x
−x
+ e (e + e ) −
− e (e + e ) dx
2
2
2(1 + e−2x ) dx
−1
]4
[
e−2x
= 2 x+
(−2) −1
[
]
[
]
e−8
e2
= 2 4−
− 2 −1 −
(2)
(2)
2
−8
= 10 + e − e
5. See Figures 5 and 6 for the illustrations:
(a) 18◦ = 18◦ ×
π
180◦
(b) 165◦ = 165◦ ×
(c)
(d)
(e)
=
π
180◦
◦
π
10
rad.
11π
rad.
12
π
3π
−270◦ = (−270 ) × 180
◦ = − 2
π
31π
930◦ = 930◦ × 180
rad.
◦ =
6
7π
π
−420◦ = (−420◦ ) × 180
◦ = − 3
=
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
rad.
rad.
0.4
0.3
0.2
11π/12
0.1
π/10
0
0
0
−0.1
−0.1
−0.1
−0.2
−0.2
−0.2
−0.3
−0.3
−3π/2
−0.4
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
−0.4
−0.4
−0.3
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
−0.4
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
Figure 5: Problems 5(a,b,c)
6. (a) We know that sin(π/6) = 12 . From the symmetry in the sine graph (Figure 7), we
infer that sin(θ) = −1
when θ = π + π/6 = 7π/6 and θ = 2π − π/6 = 11π/12.
2
(b) We know that cos(π/4) = √12 . From the symmetry in the cosine graph (Figure
8), we infer furthermore that cos(θ) = √12 when θ = 2π − π/4 = 7π/8.
√
3
(c) We know that cos(π/6) =
. From the symmetry in the cosine graph (Figure 8),
√ 2
− 3
we infer that cos(θ) = 2 when θ = π − π/6 = 5π/6 and θ = π + π/6 = 7π/6.
7
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
31π/6
−7π/3
0
0
−0.1
−0.1
−0.2
−0.2
−0.3
−0.3
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
−0.4
−0.4
0.4
−0.3
−0.2
−0.1
0
0.1
Figure 6: Problems 5(d,e)
1
0.8
0.6
y = 1/2
0.4
y
0.2
0
y = sin(x)
−0.2
−0.4
y = −1/(21/2)
−0.6
−0.8
−1
0
1
2
3
4
5
6
x
Figure 7: Problems 6(a,d)
1
y = 1/(21/2)
0.8
0.6
0.4
0.2
y
−0.4
−0.4
0
y = cos(x)
−0.2
−0.4
−0.6
y = −31/2/2
−0.8
−1
0
1
2
3
4
5
x
Figure 8: Problems 6(b,c)
8
6
0.2
0.3
0.4
(d) We know that sin(π/4) = √12 . From the symmetry in the sine graph (Figure 7),
−1
we infer that sin(θ) = √
when θ = π + π/4 = 5π/4 and θ = 2π − π/4 = 7π/4.
2
7. (a)
)
d (
sin(3x2 − 1) = cos(3x2 − 1)(6x)
dx
(b)
)
d ( 2
x + cos2 (2x) = 2x + 2 cos(2x)(−2 sin(2x))
dx
= 2x − 4 sin(2x) cos(2x)
(c)
)
d ( sin(x)
e
cos(x) = esin(x) (− sin(x)) + (cos(x))esin(x) (cos(x))
dx
= esin(x) (cos2 (x) − sin(x))
(d)
(
∫
sin
x+5
3
)
(
)(
)
x+5
1
dx = − cos
3
1/3
)
(
x+5
= −3 cos
3
(e)
∫
[
π/3
2
x + cos (2x + 1) dx =
0
=
=
=
=
=
]π/3
x3 1
+ sin (2x + 1)
3
2
0
[
] [
]
3
(π/3)
1
1
+ sin (2(π/3) + 1) − 0 + sin(1)
3
2
2
( (
)
)
3
π
1
2π
+
sin
+ 1 − sin(1)
81 2
3
( )
)
( ( )
π3 1
2π
2π
+
sin
cos(1) + cos
sin(1) − sin(1)
81 2
3
3
)
(√
π3 1
−1
3
+
cos(1) +
sin(1) − sin(1)
81 2
2
2
)
π 3 1 (√
+
3 cos(1) − 3 sin(1)
81 4
9
(f)
∫
π
3 sin
π/2
(x)
2
(x)
]π
−3(2) cos
− 4 sin(x)
2
π/2
[
(π )
] [
(π )
]
− 4 sin(π) − −6 cos
− 4 sin(π/2)
= −6 cos
2
[ (
)
]4
1
− 4(1)
= [−6(0) − 4(0)] − −6 √
2
√
= 3 2+4
− 4 cos(x) dx =
[
10