Math 217: Some Assignment 5 Solutions
14.8 # 10: Let f (x, y, z) = x2 y 2 z 2 and g(x, y, z) = x2 + y 2 + z 2 . So then ∇f =
h2xy 2 z 2 , 2yz 2 z 2 , 2zx2 y 2 i, ∇g = h2x, 2y, 2zi and the Lagrange multipier equation ∇f =
λ∇g in components is xy 2 z 2 = λx, yx2 z 2 = λy, zx2 y 2 = λz . Multiplying the first by x,
second by y, and third by z yields λx2 = λy 2 = λz 2 = x2 y 2 z 2 . If λ = 0, then at least one
of x, y, and z must be zero, in which case f (x, y, z) = 0. Otherwise, x2 = y 2 = z 2 and by
the constraint g = 1, x2 = y 2 = z 2 = 1/3, so f (x, y, z) = 1/27. Thus the minimum value
is 0, and the maximum is 1/27.
# 14: Let f (x1 , x2 , · · · , xn ) = x1 +x2 +· · ·+xn and g(x1 , x2 , · · · , xn ) = x21 +x22 +· · ·+x2n .
Then ∇f = h1, 1, · · · , 1i and ∇g = 2hx1 , x2 , · · · , xn i. So the Lagrange multiplier equation
∇f = λ∇g, in components, gives 1 = 2λxj , for j = 1, 2, . . . , n. So λ 6= 0, and x1 =
2
2
2
2
x2 = · · · = xn = 1/(2λ). From the constraint we
√ have 1 = x1 + x1 + · · · + x1 = nx1 . So
2
2
2
x1 = x2 = · · · = xn√= 1/n, and
√ so xj = ±1/
√ n for
√ j = 1, . . . , n. Thus the minimum
value√of f is f (−1/
n, ·√
· · , −1/√ n) = −n/ n = − n and the maximum value of f is
√
f (1/ n, · · · , 1/ n) = n/ n = n.
# 20: The gradient of f (x, y) = 2x2 + 3y 2 − 4x − 5 is ∇f = h4x − 4, 6yi so its only
critical point is (1, 0) which is inside the disk x2 +y 2 ≤ 16. The value there is f (1, 0) = −7.
To find the extrema on the boundary circle g(x, y) = x2 + y 2 = 16 we can use Lagrange
multipliers: ∇f = λ∇g implies 4x − 4 = 2λx, 6y = 2λy. So either y = 0, in which case
(from the
constraint)
√ constraint) x = ±4, or else λ = 3 and so x = −2 and thus (by the √
y = ± 12. The values in question are f (4, 0) = 11, f (−4, 0) = 43, f (−2, ± 12) = 47.
Thus the max is 47 and the min is −7.
15.1 # 4(b): with f (x, y) = 1 + x2 + 3y, we have f (5/4, 3/4) = 77/16 f (5/4, 9/4) =
149/16 f (7/4, 3/4) = 101/16 f (7/4, 9/4) = 173/16 and so the midpoint rule approximation
(with n = m = 2) is
Z Z
77 + 149 + 101 + 173
1
3
3
375
2
(1+x +3y)dA ≈
= 500
=
≈ 23.4
16
2
2
64
16
[1,2]×[0,3]
(True value: 47/2 = 23.5.)
#12: The integral in question is the volume over the rectangle [0, 5] × [0, 3], under the
plane z = 5 − x. This is a triangular ‘cylinder’ of cross-sectional area (1/2)(5)(5) = 25/2
and thickness 3 which has volume 75/2.
15.2 # 10:
Z 1Z 3
Z
x+3y
e
dx dy =
0
0
1
ex+3y |x=3
x=0 dy
Z
=
0
0
#16:
Z Z
2
Z
2
Z
(y + x/y )dA =
[0,2]×[1,2]
1
1
1
(e3+3y − e3y )dy = (e3 − 1) e3y |10 = (e3 − 1)2
3
3
2
2
Z
(y + x/y )dx dy =
1
0
1
1
2
(yx + x2 /(2y 2 ))|x=2
x=0 dy
2
Z
(2y + 2/y 2 )dy = (y 2 − 2/y)|21 = 4
=
1
# 26:
Z Z
2
Z
2
1
3
(3y − x + 2)dA =
V =
2
(y − x y +
Z
2y)|21
−1
[−1,1]×[1,2]
1
dx =
(7 − x2 + 2) dx
−1
= (9x − x3 /3)|1−1 = 52/3
# 36: the average value in question is
1
4
Z Z
e
[0,4]×[0,1]
y
√
x+
ey dA
1
=
4
Z
0
4
2
1
(x+ey )3/2 |y=1
y=0 dx =
3
6
Z
4
3/2
(x + e)
− (x + 1)
0
1 12
5/2
5/2
4
5/2
5/2
5/2
(x + e) − (x + 1)
|0 =
(4 + e) − 5 − e + 1
=
65
15
Oct. 28
2
3/2
dx