**Unformatted text preview: **Math for Economics II
New York University
Exam 1, Version A, Fall 2015 Name: Recitation Section: Read all of the following information before starting the exam:
• Show all work, clearly and in order, to get full credit. I reserve the right to take oﬀ points if
I cannot see how you arrived at your answer (even if your ﬁnal answer is correct).
• The exam is closed book. You are not allowed to use a calculator nor consult any notes while
taking the exam.
• This test has 3 parts and is worth 320 points. The ﬁrst part consists of 5 multiple choice
questions, the second part consists of 5 true/false questions and the last part consists of 4 free
response problems. It is your responsibility to make sure that you have all of the pages!
• The exam is 75 minutes long. Good luck! SCORES
Multiple Choice (50 pts)
TRUE-FALSE (50 pts)
Problem 1 (45 pts)
Problem 2 (60 pts)
Problem 3 (55 pts)
Problem 4 (60 pts)
TOTAL (320 pts) Multiple choice (50 points). This part consists of 5 multiple choice problems. You only
need to circle the answer, and no partial credit will be awarded.
1. Let f (x, y) = √ x − y − 1. The domain of f is (a) {(x, y) : y > x − 1}
(b) {(x, y) : y < x − 1}
(c) {(x, y) : y ≥ x − 1}
(d) {(x, y) : y ≤ x − 1}
(e) None of the above
Answer: (d)
2. Let f (x, y) = ln(2x − y). The equation of the tangent plane to the graph of z = f (x, y) at
the point (1, 1, 0) is given by
(a) z = 2x − y + 3
(b) z = 2
(x
2x−y − 1) − 2
(y
2x−y − 1) (c) z = 2x − y − 1
(d) z = x − 2y
(e) None of the above
Answer: (c)
3. Let f (x, y) =
(a) df = √ 2x + y. The total diﬀerential of f (x, y) at the point (1, 2) is given by √ 1
dx
2x+y + √1
dy
2 2x+y (b) df = 1 dx + 1 dy
2
4
(c) df = √ 2
dx
2x+y + √ 1
dy
2x+y (d) df = 1 dx + 1 dy
4
2
(e) None of the above
Answer: (b)
4. In the graph below, the grid lines are one unit apart. Which of the following statements is
true? value on the constraint x + 2y = 6, x ≥ 0, y ≥ 0
occurs at one of the endpoints.
15-tfw24
ins15-3tct1-2 subject to the constraint x − y = 0. 61. The point (2, −1) is a local minimum of f (x, y) =
x2 + y 2 subject to the constraint x + 2y = 0. For Problems 56–57, use Figure 15.39. The grid lines are one
15-tfw25 62. If grad f (a, b) and grad g(a, b) point in opposite direcunit apart.
tions, then (a, b) is a local minimum of f (x, y) cony
strained by g(x, y) = c.
ins15-3tfw26-33 g=c
f =
3 f =
5
f =
4 f =
2 In Problems 63–70, suppose that M and m are the maximum and minimum values of f (x, y) subject to the constraint
g(x, y) = c and that (a, b) satisﬁes g(a, b) = c. Decide
whether the statements are true or false. Give an explanation
for your answer. 15-tfw26 f =0 ins15-3tct1-2fig ! x 64. If f (a, b) = M , then f (a, b) = λg(a, b) for some value
of λ. 15-tfw28 f =
1 63. If f (a, b) = M , then fx (a, b) = fy (a, b) = 0. 15-tfw27 65. If grad f (a, b) = λ grad g(a, b), then f (a, b) = M or
f (a, b) = m. 15-tfw29 66. If f (a, b) = M and fx (a, b)/fy (a, b) = 5, then
gx (a, b)/gy (a, b) = 5. Figure 15.39 15-tfw30 67. If f (a, b) = m and g (a, b) = 0, then f (a, b) = 0.
x
x
785
15-tfw31 68. Increasing the value of c increases the value of M .
56. Find the maximum and minimum values of f on g = c.
(a) points do they occur?
At whichThe maximum value of f on the line g Suppose that f (a,it occurs at that grad f (a, (4, =
15-tfw32 69. = c is 2 and b) = M and the point b) 2) 14.4 GRADIENTS AND DIRECTIONAL DERIVATIVES IN THE PLANE
15-3tct1 14.4 15-3tct2 57. Find the maximum and minimum values of fon the tri(b) The minimum value of f on the line
angular region below g = c in the ﬁrst quadrant. g 3 gradis 2 b). Then occurs at the pointby 0.02
= c g(a, and it increasing the value of c (4, 2)
increases the value of M by about 0.06. 15-tfw33 70. below the (a, b)
(c) The minimum value of f on the regionSuppose that f line g==m and2that
c is grad f (a, b) =
ins15-3tfw1-25
14-4w27
Are the f = (2x + 3ey )i + 3xey j
on. Assume 29. grad statements in Problems 58–62 true or false? Give rea3 grad g(a, b). Then increasing the value of c by 0.02
sons for(d) answer.
your The maximum value of f on the region below the line g = c is 2
the function
decreases the value of m by about 0.06.
ins14-4w28-29
In Exercises 30–31, ﬁnd grad f from the differential. (e) None of the above 14-4w28 + 100L 14-4w29 ins14-4w30-35 r2 h
L 0.7 x (see also Ready Reference at the end of the book) x 31. df = (x + 1)ye (b)+ xe dy
Answer: dx
• Critical Points
In Exercises 32–37, contour diagram ofof fbelow to
5.Deﬁnitions of use the extrema, diagram points, saddle
Use the local contour critical f in Figure 14.34 to decide if the speciﬁed directional derivative is
points, ﬁnding critical points algebraically, behavior
positive, the pointapproximately zero.discriminant, secondof negative, or (−2, 2):points,
contours near critical
derivative test to classify critical points.
y
• Unconstrained Optimization
3
Deﬁnitions of global extrema, method of least squares, n(x2 + y 2 ) closed and bounded regions, existence of global extrema.
• Constrained Optimization
determine the signs of the directional derivatives at
Objective function and constraint, deﬁnitions of extrema
subject to a constraint, geometric interpretation of Lagrange multiplier method, solving Lagrange multiplier
problems algebraically, inequality constraints, meaning
of the Lagrange multiplier λ, the Lagrangian function. 8 2 x/y) 6 K 0.3 CHAPTER + 10ydy
30. df = 2xdxSUMMARY 4 2 REVIEW1EXERCISES AND PROBLEMS FOR CHAPTER FIFTEEN
x Exercises
2 −1 6 4 15-1dhhwnp002
For Exercises 1–6, ﬁnd the critical points of the given func1. f (x, y) = x2 + 2xy − y 2 − 4x − 8y + 9
−2
tion and classify them as local maxima, local minima, saddle
15-miscw1 2. f (x, y) = 2xy 2 − x2 − 2y 2 + 1
points, or none of these.
−3
15-1w9 3. f (x, y) = x3 + y 2 − 3x2 + 10y + 6
ins14-4w30-35fig
−3 −2 −1
1
2
3 ins15-miscw1-5 8 ! Figure 14.34
14-4w30 32. At point (−2, 2), in direction i . 14-4w31 (a) Positive in the
33. At point (0, −2), in direction j . directions 14-4w34
14-4w35
14-4w32 ve fu (1, 2)
14-4w33
ins14-4w36-43 x − 4y in(2x − y)
14-4w38 34. At point (0, −2), in direction i + 2j .
(b) Positive in the direction of both i and j of i, negative in the direction of j 35. At point (0, −2), in direction i − 2j . (c) Negative in the direction of i, positive in the direction of j 36. At point (−1, 1), in direction i + j . (d) Negative in the directions of both i and j 37. At point (−1, 1), in direction −i + j . (e) Cannot determine from the data. In Exercises 38–45, use the contour diagram of f in Figure 14.34 to ﬁnd the approximate direction of the gradient
vector atAnswer: (c)
the given point.
14-4w39
38. (−2, 0) 39. (0,14-4w36 40. (2,14-4w37 41. (0, 2)
−2)
0) e directional
14-4w43 43. (−2, −2) 44. (2,14-4w41 45. (2, −2)
14-4w40
14-4w42 42. (−2, 2)
2)
of v . he gradient. True or false (50 points). This part consists of 5 true/false questions. For each statement below, circle TRUE if the statement is always true. Otherwise, circle FALSE.
You do not need to show your work.
1. If fx (a, b) = 0 and fy (a, b) = 0, then the directional derivative fu (a, b) is zero for any u. TRUE FALSE 2. If f (x, y) has a local maximum at (a, b) subject to the constraint g(x, y) = c, then g(a, b) = c. TRUE FALSE 3. If f (x, y) has a local maximum at the point (a, b) subject to the constraint g(x, y) = c, then
fx (a, b) = 0 and fy (a, b) = 0. TRUE FALSE 4. The function f (x, y) = x − y has no global minimum subject to the constraint x + y = 0. TRUE FALSE TRUE FALSE 5. If u and v are vectors in R3 , then |u + v| ≤ |u| + |v|. Problems 1-4 are free response problems. Put your work/explanations in the space
below the problem. Read and follow the instructions of every problem. Show all your
work for purposes of partial credit. Full credit may not be given for an answer alone.
Justify your answers.
1. Find the following partial derivatives:
(a) (15 pts) fx if f (x, y) =
fx =
= x ln x
.
x2 +y 2 Solution: (x ln x)x (x2 + y 2 ) − x ln x(x2 + y 2 )x
(x2 + y 2 ) − x ln x · 2x
=
=
(x2 + y 2 )2
(x2 + y 2 )2
(ln x + 1)(x2 + y 2 ) − x ln x · 2x
(−x2 + y 2 ) ln x + (x2 + y 2 )
=
.
(x2 + y 2 )2
(x2 + y 2 )2 (b) (15 pts) WA if W (A, B, C) = AA B A C A . Solution: we use logarithmic diﬀerentiation:
ln W = A ln A + A ln B + A ln C, WA
= ln A + ln B + ln C + 1,
W WA = AA B A C A (ln A + ln B + ln C + 1).
(c) (15 pts) zy if x + y + z + x2 + y 2 + z 2 = 0. Solution: we use implicit diﬀerentiation:
(x + y + z + x2 + y 2 + z 2 )y = 0,
1 + zy + 2y + 2zzy = 0,
zy (1 + 2z) = −(1 + 2y),
zy = − 1 + 2y
.
1 + 2z 2. Let f (x, y) = xye−y .
(a) (30 pts) Find the instantaneous rate of change of f at the point (3, 2) in the direction of
the point (0, 6).
Solution: we ﬁnd the gradient:
fx = ye−y , fy = xe−y − xye−y , f (3, 2) = 2e−2 , −3e−2 . A vector from (3, 2) to (0, 6) is −3, 4 , a unit vector in the same direction is u =
−3/5, 4/5 , so
Du f (3, 2) = 2e−2 , −3e−2 · −3/5, 4/5 = − 18e−2
.
5 (b) (10 pts) What is the direction of the maximum instantaneous rate of change at (3, 2)?
Solution: the direction of the maximum rate of change is given by the gradient, or 2, −3 .
(c) (10 pts) What is the value of the maximum instantaneous rate of change at (3, 2)?
Solution:√
the value of the maximum rate of change is the magnitude of the gradient,
which is 13e−2 .
(d) (10 pts) Give a nonzero vector in the direction of which the instantaneous rate of change
at (3, 2) is zero.
Solution: we need to pick a vector orthogonal to the gradient, for example 3, 2 . 3. Given x units of capital and y units of labor, a company produces
f (x, y) = 10x1/4 y 1/3 units of its product. One unit of capital costs $3 and one unit of labor costs $8.
(a) (40 points) Suppose the company’s budget is $112. Use Lagrange multipliers to ﬁnd
the values of x and y which maximize production, and ﬁnd the maximum production
level. You may assume that the max occurs where the Lagrange multipliers condition is
satisﬁed.
Solution: the constraint is g(x, y) = 3x + 8y = 112, so we set up the Lagrange equations:
1
10 4 x−3/4 y 1/3 = 3λ
10 1 x1/4 y −2/3 = 8λ
3
3x + 8y = 112 → 10
λ = 3·4 x−3/4 y 1/3
10
λ = 3·8 x1/4 y −2/3
3x + 8y = 112 → 10
λ = 3·4 x−3/4 y 1/3
10
10 −3/4 1/3
x
y = 3·8 x1/4 y −2/3
3·4
3x + 8y = 112 From the second equation we get
10 −3/4 1/3
10 1/4 −2/3
x
y =
x y
3·4
3·8 → 10
10
y=
x
3·4
3·8 → x = 2y, and plugging this into the third equation gives us x = 16 and y = 8. This is where the
maximum occurs, and the maximum value is
f (16, 8) = 10 · 161/4 81/3 = 40.
(b) (20 points) Without re-doing the optimization problem in part (a), estimate how much
the new maximum value of f would change if the budget were increased from $112 to
$115.
Solution: at the maximum point, the value of λ is
λ= 5
10 −3/4 1/4 5 2
16
8 =
= ,
12
68
24 this is the rate of change of the maximum value if we change the budget by one dollar.
We’re changing the budget by three dollars, so the maximum value will increase by 5/8. 4. (60 pts) Find the extreme values of f (x, y) = x2 +y 2 −2x subject to the constraint x2 +y 2 ≤ 1.
(a) (20 pts) Find the critical points of f in the open region x2 + y 2 < 1. Solution: we ﬁnd
where f = 0:
fx = 2x − 2 = 0, fy = 2y = 0, → (x, y) = (1, 0). However, the point (1, 0) is not in the open region x2 + y 2 < 1, so we won’t need it.
(b) (30 pts) Use Lagrange multipliers to ﬁnd the critical points of f on x2 + y 2 = 1.
Solution: let g(x, y) = x2 + y 2 , then the Lagrange equations are
2x − 2 = 2λx
2y = 2λy .
x2 + y 2 = 1
From the second equation, we get that either λ = 1 or y = 0. If λ = 1, then the ﬁrst
equation becomes 2x−2 = 2x, which is a contradiction. If y = 0, then the third equation
implies that x = 1 or x = −1, and the ﬁrst equation can be solved for λ = (x − 1)/x.
Hence there are two critical points (x, y, λ) = (1, 0, 0) and (−1, 0, −2).
(c) (10 pts) Use these to nd the absolute maximum and absolute minimum values of f on
the domain x2 + y 2 ≤ 1.
Solution: we compare the values of f at (1, 0) and (−1, 0), and we see that f (1, 0) = −1
is the absolute minimum and f (−1, 0) = 3 is the absolute maximum.
• The absolute maximum occurs at the point/s
.
is and the maximum of f (x, y) • The absolute minimum occurs at the point/s
is
. and the minimum of f (x, y) This page is intentionally left blank for computations ...

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