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Calculus: Early Transcendental Functions
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Chapter 2 / Exercise 129
Calculus: Early Transcendental Functions
Edwards/Larson
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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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