MATHEMATICS 226, FALL 2014, PROBLEM SET 5
Solutions1
Section 13.1, Question 25: Not marked
Let (x, y, z) be the coordinates of the corner of the box in the octant x, y, z ≥
0. We want to maximize the function V (x, y, z) = 8xyz subject to the
x2 y 2 z 2
constraint g(x, y, z) = 2 + 2 + 2 = 1. We have
a
b
c
∇f = h8yz, 8xz, 8xyi, ∇g = h
2x 2y 2z
, , i
a2 b2 c2
We use Lagrange multipliers:
8yz = λ
2y
2z
2x
,
8xz
=
λ
,
8xy
=
a2
b2
c2
4a2 yz
. (If any one of x, y, z is 0,
x
then V = 0, and that’s clearly not the maximum value we are looking for.)
Plugging this into the second equation, we get
From the first equation, we have λ =
8xz =
4a2 yz 2y 8x2 z
8y 2 z x2
y2
,
=
,
=
x b2
a2
b2
a2
b2
y2
z2
By symmetry, we also have 2 = 2 . Using also the equation of the ellipsoid,
b
c
we get that
x2 y 2 z 2
x2
a
b
c
√
√
√
+
+
=
3
=
1,
x
=
,
y
=
,
z
=
.
a2
b2
c2
a2
3
3
3
Section 13.1, Question 27: 10 marks: 4 for implicit differentiation, 2 for
plugging in zx = zy = 0, 4 for solving the equations.
We use implicit differentiation to differentiate the given equation in x and y:
2zx−x2
e
∂z ∂z
2zy+y 2
2z + 2x
− 2x − 3e
2y
= 0,
∂x
∂x
1
c Laba. Not to be copied, used, or revised without explicit written permission from
I.
the copyright owner.
1
∂z ∂z
2
2
e2zx−x 2x
− 3e2zy+y 2y
+ 2z + 2y = 0.
∂y
∂y
At a critical point, we have
∂z
∂x
=
∂z
∂y
= 0, so that
2
e2zx−x (2z − 2x) = 0, z = x,
2
−3e2zy+y (2z + 2y) = 0, z = −y
Substituting z = x and y = −x into the given equation, we get
e2x
2 −x2
− 3e−2x
2 +x2
2
2
= ex − 3e−x = 2
2
Let u = ex , then u−3u−1 = 2, u2 −3 = 2u, u2 −2u−3 = 0, u = 3 or u = −1.
2
The second value√is not possible since u = ex > 0, so we√must have
√ u = 3,
x2 √
= ln 3,√x = ± ln 3. The critical points are (x, y) = ( ln 3, − ln 3) and
(− ln 3, ln 3).
Section 13.2, Question 11: 10 marks: 2 for the critical points inside the
ball, 2 for setting up Lagrange equations, 6 for solving them.
Let f (x, y, z) = xy 2 + yz 2 . We first look for critical points: ∇f = hy 2 , 2xy +
z 2 , 2yzi, so that at a critical point we must have y = 0, z = 0. There
is no equation for x, so we have a family of critical points (x, 0, 0) with
f (x, 0, 0) = 0.
Next, we look for the extreme values on the surface of the ball. We use
Lagrange multipliers. Let g(x, y, z) = x2 + y 2 + z 2 , then ∇g = h2x, 2y, 2zi.
We get equations
y 2 = 2λx, 2xy + z 2 = 2λy, 2yz = 2λz.
From the last equation, z = 0 or λ = y.
• If z = 0, then from the second equation 2xy = 2λy, so that y = 0
(a case we’ve covered already) or λ = x. In the latter case, from the
first equation y 2 = 2x2 , and plugging
√ it into the equation of the sphere,
2
2
2
2
x + y + z = 3x = 1, x = ±1/ 3. We get four critical points
r 1 r 2 r1 r2 2
2
f
,±
,0 = √ , f −
,±
,0 = − √ .
3
3
3
3
3 3
3 3
2
• If λ = y, then from the first two equations we have y 2 = 2xy and
2xy + z 2 = 2y 2 , so that y 2 + z 2 = 2y 2 , y 2 = z 2 . Also, from y 2 = 2xy we
have y = 0 or y = 2x. If y = 0, then also z = 0 and we’ve covered this
case already. If y = 2x, then x2 + y 2 + z 2 = y 2 + 4x2 + 4x2 = 9x2 = 1,
so that x = ±1/3. We get four critical points:
1 2 2 4
1 2 2
4
f , ,±
= , f − ,− ,±
=− .
3 3 3
9
3 3 3
9
Since 49 > 3√2 3 , the maximum and minimum values are 4/9 and −4/9, at the
above points.
Section 13.3, Question 7: 10 marks: 2 for setting up the correct constraint, 2 for setting up the Lagrange equations, 6 for solving them.
1
4
1
We need to minimize f (a, b, c) = abc with the constraint 2 + 2 + 2 = 1.
a
b
c
Note that we can’t have a, b or c = 0 since then the equation of the ellipsoid
would make no sense, and we may as well assume that a, b, c > 0. Let
g(a, b, c) = a−2 + 4b−2 + c−2 , then
∇f = hbc, ac, abi, ∇g = h−2a−3 , −8b−3 , −2c−3 i
so we get a system of equations
bc = −2λa−3 , ac = −8λb−3 , ab = −2λc−3
or equivalently
a3 bc = −2λ, ab3 c = −8λ, abc3 = −2λ
Dividing by abc (since a, b, c 6= 0,
a2 = −2λ, b2 = −8λ, c2 = −2λ
so that c2 = a2 and b2 = 4a2 . Plugging this into the constraint, we get
√
√
1
4
1
3
+
+
=
=
1,
so
that
a
=
c
=
3,
b
=
2
3.
a2 b2 c2
a2
Section 13.3, Question 15: Not graded.
Let g1 (x, y, z) = x2 + y 2 and g2 (x, y, z) = x + y + z, then
∇f = h0, 0, −1i, ∇g1 = h2x, 2y, 0i, ∇g2 = h1, 1, 1i
3
Using Lagrange multipliers, we get equations
0 = 2λ1 x + λ2 , 0 = 2λ1 y + λ2 , −1 = λ2 .
From the first two equations, we have 2λ1 x = 2λ2 y, so λ1 = 0 or x = y.
• If λ1 = 0, then from the first equation we get λ2 = 0, but that contradicts the third equation. No solutions here.
• If x = y, then x2 + y 2 = 2x2 = 8 yields x2 = 4, x = ±2. We also
have z = 1 − x − y = 1 − 2x. We get two critical points (2, 2, −3) and
(−2, −2, 5), with
f (2, 2, −3) = 7, f (−2, −2, 5) = −1
These are the maximum and minimum values, respectively.
Section 13.3, Question 25: 10 marks: 3 for each implicit differentiation
(f and g), 4 for putting it together.
Let y = h(x) be the function implicitly defined by g(x, y) = g(a, b), i.e.
g(x, h(x)) = g(a, b), h(a) = b. Differentiating in x, we get
0=
d
g(x, h(x)) = g1 (x, h(x)) + g2 (x, h(x))h0 (x).
dx
Assuming that g2 (a, b) 6= 0, we get
h0 (a) = −
g1 (a, b)
.
g2 (a, b)
If f (x, h(x) has a critical point at x = a, it has a critical point there:
d
0
f (x, h(x))
= (f1 (x, h(x)) + f2 (x, h(x))h (x))
0=
dx
x=a
x=a
= f1 (a, b) − f2 (a, b)
g1 (a, b)
.
g2 (a, b)
f2 (a, b)
, then from the last equation f1 (a, b) = λg1 (a, b), so that
g2 (a, b)
∇f (a, b) = λ∇g(a, b) and the Lagrange multiplier condition is satisfied.
Let λ =
4