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Math 241, Exam 2. 10/26/11.
Name:
• Read problems carefully. Show all work. No notes, calculator, or text.
• The exam is roughly 15 percent of the total grade. There are 100 points total.
• Write your full name in the upper right corner of page 1.
• Do problem 1 on p. 1, problem 2 on p. 2,...(or at least present your solutions in numerical order).
• Circle or otherwise clearly identify your final answer.
x4 + 3y 3
. Does lim f (x, y) exist? If it exists,
(x,y)→(0,0)
2x3 − y 3
give its value. If it does not exist, briefly explain how you reached your conclusion.
1. (10 points): Suppose that f (x, y) =
Solution: Approach along x = 0 and along y = 0.
• Along x = 0: We have
3y 3
lim f (0, y) = lim
= −3.
y→0
y→0 −y 3
• Along y = 0: We have
x4
= 0.
x→0 2x3
lim f (x, 0) = lim
x→0
Since these limits disagree, we see that
x4 + 3y 3
does not exist.
(x,y)→(0,0) 2x3 − y 3
lim
2. (10 points): Find an equation of the form ax + by + cz = d for the plane tangent to the
surface F (x, y, z) = y − x2 + z 2 = 0 at the point (4, 7, 3).
Solution: We have
∇F (x, y, z) = h−2x, 1, 2zi, ∇F (4, 7, 3) = h−8, 1, 6i.
It follows that the tangent plane has equation
−8(x−4)+1(y−7)+6(z−3) = 0 ⇐⇒ −8x+y+6z = −32+7+18 = −7 ⇐⇒ 8x−y−6z = 7.
3. (20 points):
(a) (12 points): Suppose that
u = z sin x + zy 3 − y 2 , x = et , y = ln(t + 1), z = 3t5 + 2t + 1.
Use the Chain rule to compute du/dt. You do not need to put the answer in terms of t
alone; you may use the variables x, y, and z as well.
Solution: We have
du
∂u dx ∂u dy ∂u dz
1
t
2
=
+
+
= z cos x·e +(3zy −2y)·
+(sin x+y 3 )(15t4 +2).
dt
∂x dt ∂y dt ∂z dt
t+1
(b) (8 points): Suppose that
z = F (x, y, t), x = G(t), y = H(s, t).
Use partial derivatives of F , G, and H to write down the form of the chain rule that
you would use to compute ∂z/∂t. (Use a tree diagram to show variable dependencies.)
Solution:
∂F dG ∂F ∂H ∂F
∂z
=
+
+
.
∂t
∂x dt
∂y ∂t
∂t
4. (20 points): Let f (x, y, z) = xy + z 2 .
(a) (13 points): Find the directional derivative of f (x, y, z) at (1, 1, 1) in the direction
of (5, −3, 3).
Solution: We have
h4, −4, 2i
2 2 1
~u = √
=
,− ,
, ∇f (x, y, z) = hy, x, 2zi, ∇f (1, 1, 1) = h1, 1, 2i.
3 3 3
36
It follows that
D~u f (1, 1, 1) = h1, 1, 2i ·
2 2 1
,− ,
3 3 3
=
2
2 2 2
− + = .
3 3 3
3
(b) (7 points): Give a unit vector in the direction in which f decreases most rapidly at
(1, 1, 1).
Solution: We have
∇f (1, 1, 1)
h1, 1, 2i
−
=− √
=
k∇f (1, 1, 1)k
6
1
1
2
−√ , −√ , −√
6
6
6
.
5. (25 points): Let f (x, y) = x2 − 6xy + y 2 + 4x + 4y.
(a) (12 points): Use the second derivative test to locate all local maxima, minima, and
saddle points of f (x, y).
Solution: We solve the system:
fx (x, y) = 2x − 6y + 4 = 0 ⇐⇒ x − 3y + 2 = 0 ⇐⇒ x = 3y − 2
fy (x, y) = 2y − 6x + 4 = 0 ⇐⇒ y − 3x + 2 = 0 ⇐⇒ y = 3x − 2.
We substitute the first equation in the second to obtain
y = 3(3y − 2) − 2 = 9y − 8 ⇐⇒ y = 1, x = 1.
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It follows that (1, 1) is the unique critical point. To apply the second derivative test, we
compute
fxx (x, y) = 2, fyy (x, y) = 2, fxy (x, y) = fyx (x, y) = −6,
2 −6
= 4 − 36 = −32 < 0.
D = −6 2 We conclude that (1, 1) is a saddle point.
(b) (13 points): Let D = {(x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 5}. Find the absolute
maximum and minimum of f (x, y) on D. Justify your conclusions.
Solution: It suffices to analyze f (x, y) on the boundary of D.
• x = 0: f (0, y) = u1 (y) = y 2 + 4y; u01 (y) = 2y + 4 = 0 ⇐⇒ y = −2;
(0, −2) 6∈ D.
• y = 0: f (x, 0) = u2 (x) = x2 + 4x; (−2, 0) 6∈ D.
• x = 2: f (2, y) = u3 (y) = 4 − 12y + y 2 + 8 + 4y = y 2 − 8y + 12; u03 (y) =
2y − 8 = 0 ⇐⇒ y = 4; (2, 4) ∈ D.
• y = 5; f (x, 2) = u4 (x) = x2 − 26x + 45; u04 (x) = 2x − 26 = 0 ⇐⇒ x = 13;
(13, 5) 6∈ D.
We note that the critical point (1, 1) in the interior of D is a saddle, so f does not have
an extremum there. It remains to evaluate f at (2, 4) and at the corners: (0, 0), (0, 5),
(2, 0), (2, 5). We have:
f (2, 4) = −4, f (0, 0) = 0, f (0, 5) = 45, f (2, 0) = 12, f (2, 5) = −3.
Therefore, the absolute minimum is at (2, 4) while the absolute maximum is at (0, 5).
6. (15 points): Use Lagrange multipliers to find the point(s) (x, y, z) which maximize and
minimize f (x, y, z) = x + 2y + 8z subject to the constraint g(x, y, z) = x2 + y 2 + 2z 2 = 37.
Solution: We compute
∇f (x, y, z) = h1, 2, 8i , ∇g(x, y, z) = h2x, 2y, 4zi .
The equation ∇f (x, y, z) = λ∇g(x, y, z) gives
1
1
2
, 2 = 2λy ⇐⇒ y = , 8 = 4λz ⇐⇒ z = .
2λ
λ
λ
We substitute these values in g(x, y, z) = 37 to obtain
2 2
2
1
1
2
1 + 4 + 32
37
g(x, y, z) =
+
+2
=
= 2 = 37 ⇐⇒ λ = ±1/2.
2
2λ
λ
λ
4λ
4λ
1 = 2λx ⇐⇒ x =
We substitute λ in for x, y, and z to obtain points
(1, 2, 4), (−1, −2, −4).
The maximum is at f (1, 2, 4) = 37; the minimum is at f (−1, −2, −4) = −37.
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