Quiz 2B for MATH 105 SECTION 205
January 23, 2015
Given Name
Family Name
Student Number
1. (a) (1 point) Find the value of a such that the gradient of f (x, y, z) = aex + xy + sin(z) at point (0, 0, 0) is
orthogonal to a normal vector of the plane 2x − z = 1.
1
2
(a)
Answer.
Since
fx = aex + y,
fy = x,
and fz = cos(z),
then
∇f (0, 0, 0) = ha, 0, 1i.
Notice that a normal vector of the plane 2x − z = 1 can be h2, 0, −1i, by the assumption, we have
0 = ha, 0, 1i · h2, 0, −1i = 2a − 1.
Then
a=
1
2
(b) (1 point) Can you find a function h(x, y) such that ∇h(x, y) = hcos(y), sin(x)i? (Do not need to find
h(x, y), just put ‘Yes’ or ‘No’)
No
(b)
Answer.
If we can find h(x, y) such that ∇h(x, y) = hcos(y), sin(x)i, then
hx = cos(y),
and hy = sin(x).
By Clairaut’s theorem, we must have hxy = hyx . But we know that
hxy = − sin(y),
and hyx = cos(x),
contradiction.
(c) (2 points) Solve the system
x3 = xy
.
y−x=2
(c)
Answer.
(0, 2), (−1, 1) and (2, 4)
From the first equation, x3 = xy, we have x(x2 − y) = 0, then either x = 0 or y = x2 .
• When x = 0, plug x = 0 into the second equation y − x = 2, then y = 2. So we have one solution
(0, 0).
• When y = x2 , plug y = x2 into the second equation x2 − x = 2, that is, x2 − x − 2 = 0, then x = 2
or x = −1. So we have two solutions (2, 4) and (−1, 1).
In summary, we have three solutions:
(0, 2),
(−1, 1),
and
(2, 4).
2. (4 points) Use the Lagrange multipliers to find the maximum and minimum values of f (x, y) = xy subject to
x2 + y 2 − xy = 9.
Answer.
The objective function is f (x, y) = xy, and the constraint is g(x, y) := x2 + y 2 − xy − 9 = 0, so
we should solve the following system:
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
Notice that
∇f (x, y) = hy, xi,
and ∇g(x, y) = h2x − y, 2y − xi.
So we need to solve the system:
2
y = λ(2x − y)
(1)
x = λ(2y − x)
(2)
2
x + y − xy = 9
(3)
It’s easy to see that λ 6= 0, otherwise, by (1) and (2), we have x = y = 0, which does not satisfy (3). By (1)
and (2), we have
yλ(2y − x) = xλ(2x − y).
Sine λ 6= 0, then
2y 2 − xy = 2x2 − xy.
That is, y 2 = x2 , then y 2 − x2 = 0. So (y − x)(y + x) = 0, then y = x or y = −x.
When y = x, plug y = x into (3), we get x2 + x2 − x2 = 9, that is, x2 = 9. So x = 3 or x = −3. Since
y = x,then y = 3 or y = −3. So when (x, y) = (3, 3), by (1), we get 3 = λ(2 · 3 − 3), then λ = 1. When
(x, y) = (−3, −3), by (1), we get −3 = λ(−2 · 3 + 3), then λ = 1. So we have two solutions:
(x, y, λ) = (3, 3, 1),
and
(x, y, λ) = (−3, −3, 1).
√
√
2 + x2 + x2 = 9, that is, x2 = 3. So x =
When y = −x, plug y = √
−x into (3),
we
get
x
3
or
x
=
−
√
√
√
√
√
√ 3.
Since y = −x,then y = − 3 or
√y =
√ 3. So when (x, y)√= ( 3, − 3),
√ by (1),
√ we get 3 = λ(2 · 3 + 3),
then λ = 31 . When (x, y) = (− 3, 3), by (1), we get − 3 = λ(−2 · 3 − 3), then λ = 13 . So we have two
solutions:
√ √ 1
√
√ 1
(x, y, λ) = ( 3, − 3, ), and (x, y, λ) = (− 3, 3, ).
3
3
Notice that
f (3, 3) = f (−3, −3) = 9,
√
√
√ √
and f ( 3, − 3) = f (− 3, 3) = −3.
So the maximum value of f (x, y) = xy subject to x2 + y 2 − xy = 9 is 9, and the minimum value of f (x, y) = xy
subject to x2 + y 2 − xy = 9 is −3.
3. Let f (x, y) = 2x2 + 2y 2 − 6x and R := {(x, y) : x2 + y 2 ≤ 9}, then
(a) (1 point) Use polar coordinates to find the maximum and minimum values of f (x, y) on the boundary of
R.
Answer.
For (x, y) on the boundary of R, then x2 + y 2 = 9. Use polar coordinates, we have
x = 3 cos(θ),
for all 0 ≤ θ ≤ 2π.
y = 3 sin(θ),
Then
g(θ) = f (x, y)
= f (3 cos(θ), 3 sin(θ))
= 2 · 9 cos2 (θ) + 2 · 9 sin2 (θ) − 6 · 3 cos(θ)
= 18 cos2 (θ) + 18 sin2 (θ) − 18 cos(θ)
= 18 − 18 cos(θ),
for all 0 ≤ θ ≤ 2π.
So g 0 (θ) = 18 sin(θ). If g 0 (θ) = 0, that is, sin(θ) = 0. Since 0 ≤ θ ≤ 2π, then θ = 0, π, 2π.
Notice that
g(0) = g(θ) = 18 − 18 cos(0) = 0,
and g(π) = 18 − 18 cos(π) = 18 + 18 = 36.
So the maximum value of f (x, y) on the boundary of R is 36, and the minimum value of f (x, y) on the
boundary of R is 0.
(b) (1 point) Use Lagrange multipliers to find the maximum and minimum values of f (x, y) on the boundary
of R.
Answer.
Let g(x, y) = x2 + y 2 − 9, then the objective function is f (x, y) = 2x2 + 2y 2 − 6x and the
constraint is g(x, y) = x2 + y 2 − 9 = 0. So we need to solve the system:
∇f (x, y) = λ∇g(x, y),
and g(x, y) = 0.
Notice that
∇f (x, y) = h4x − 6, 4yi,
∇g(x, y) = h2x, 2yi.
So we need to solve the system:
4x − 6 = 2λx
(4)
4y = 2λy
(5)
2
x +y
If λ = 0, by (4) and (5), then x =
(6).
If λ 6= 0, by (4) and (5), we have
3
2
2
= 9
(6)
and y = 0. It’s easy to see that (x, y) = (3/2, 0) is not a solution to
2λy(4x − 6) = 2λx(4y).
Since λ 6= 0, then 2y(4x − 6) = 8xy, that is, 8xy − 12y = 8xy. So y = 0. Plug y = 0 into (6), we get
x2 = 9, then x = 3 or x = −3. If x = 3, by (4), then 6 = 6λ, that is, λ = 1. If x = −3, by (4), then
−12 = −6λ, that is, λ = 2. We get two solutions
(x, y, λ) = (3, 0, 1),
and
(x, y, λ) = (−3, 0, 2).
and
(x, y, λ) = (−3, 0, 2).
In summary, we get two solutions:
(x, y, λ) = (3, 0, 1),
Notice that
f (3, 0) = 2 · 32 − 6 · 3 = 0,
and f (−3, 0) = 2 · (−3)2 + 6 · 3 = 36.
So the maximum value of f (x, y) on the boundary of R is 36, and the minimum value of f (x, y) on the
boundary of R is 0.
(c) (1 point) Find the absolute maximum and minimum values of f (x, y) on R.
Answer.
Find critical points of f (x, y) inside R. Since ∇f (x, y) = h4x − 6, 4yi, then we have only one
critical point:
3
(x, y) = ( , 0).
2
Notice that
3
f ( , 0) = 2 ·
2
2
3
3
9
9
−6· =2· −9=−
2
2
4
2
By the result of part a, we know that the absolute maximum value of f (x, y) on R is 36, and the absolute
minimum value of f (x, y) on R is − 92 .
Your Score:
/11