Department of Computer & Mathematical Sciences University of Toronto at Scarborough MATA33H3Y: Calculus for Management II Final Examination August, 213 Examiner: A. Chow Surname (print): Given Name(s) (print): Student Number: Signature: Tutorial Code and TA (circle one): T1 Charles Tsang T2 Stephen Tang T3 Jason Zhang T5 Stephen Tang T4 Helen Wang T6 Jason Zhang Instructions: 1. This exam has 14 pages, excluding this cover page. It is your responsibility to ensure no pages are missing. 2. Answer the multiple choice questions by filling in the Scantron sheet and circling the answer. Answer all full solution questions in the spaces provided; if you need extra space, you may use the back of a page, but clearly indicate where the continuation of your work is. 3. With the exception of multiple choice questions, you must show all steps and use the methods of MATA33 to obtain full marks. 4. One standard calculator (as stated in the course outline) is allowed. No other aids allowed. 5. Time allowed: 3 hours. Question: MC 2 3 4 5 6 7 8 Total Points: 39 8 1 8 7 8 11 9 1 Score:
Part I Multiple Choice 1. Indicate the correct answer on the Scantron sheet and also circle the correct answer. A correct answer will earn you 3 points. An incorrect or unanswered question will be worth points. Justification is neither required nor rewarded. [ 1 2 3 4 (i) (3 points) Let A = [ ] 7 11 A. 3 13 [ ] 1 7 B. 9 11 [ ] 11 13 C. 14 [ ] 1 3 D. 1 4 [ ] 5 8 E. 2/3 25/2 ] and B = [ 1 2 3 4 ]. What is 3A T + 2B 1? (ii) (3 points) Let Z = 2Y + 3X. If R is the feasible region defined by R = { (x, y) R 2 : x, y x + 3 } then A. Z has both a maximum and a minimum on R. B. Z has a maximum, but no minimum on R. C. Z has a minimum, but no maximum on R. D. Z has neither a maximum nor a minimum on R. E. Z is undefined on R. Remark: This is adapted from the 212 winter exam. (iii) (3 points) The joint demand functions for products A and B are given by q A = 5 3 P B and q B = 75P A PA where P A and P B are the prices of A and B respectively. Then A. A and B are complimentary products. B. A and B are competitive products. C. A and B are both complimentary and competitive products. D. A and B are neither complimentary nor competitive products. E. None of the above. Remark: This is 17.2 Example 4. 3 P 2 B Page 1 of 14
1 4 7 (iv) (3 points) What is det 2 5 8? 3 6 9 A. B. 1 C. -1 D. 13 E. 45 Remark: This is an example from lecture, and also a smaller version of Question 7 from Assignment 4. (v) (3 points) The value of 2 x 2 xy dy dx is A. B. 1 C. 16 5 D. 16 3 E. undefined Remark: This is 17.9 #2 from Assignment 6. 2 1 7 (vi) (3 points) Let A = 3 5 and let B be a square matrix of order 3 such that 1 det B = 3. What is det ( 2A 1 B 2) A. 1 B. 2 C. 3 D. 9 E. 12 Page 2 of 14
(vii) (3 points) Suppose the equation r 2 s 2 s 2 + t 2 = t2 2 defines r as a function of the independent variables s and t. The value of r t s = 1, and t = 1 is A. B. 1 1 C. 2 D. 3 2 E. undefined Remark: This is 17.3 #18 from Assignment 5. when r = 1, 2 4 6 x 8 (viii) (3 points) Let A = 1 2 3, X = y, and B =. The matrix equation AX = B 1 2 3 z 1 has A. no solution. B. one solution. C. two solutions. D. infinitely many solutions. E. only the trivial solution. Remark: This is adapted from 6.4 #9 from Assignment 3. (ix) (3 points) Suppose a function f(x, y) has continuous second-order partial derivatives at all points (x, y) R 2 and that (a, b) is a critical point of f where a < b and b > [.157. ] ab b Suppose the Hessian matrix, i.e. the matrix of second order partial derivatives is. b 1 Then we may conclude that A. f has a relative maximum at (a, b). B. f has a relative minimum at (a, b). C. f has no relative extrema at (a, b). D. nothing can be said about the relative extrema of f from the information given. E. none of the above is true. Remark: This is adapted from the 211 winter exam. Page 3 of 14
(x) (3 points) Exactly how many of the following mathematical statements are equivalent to the statement The 3 3 matrix A is invertible? det A =. The matrix equation AX = has only the trivial solution. AB = I for some 3 3 matrix B. A is equivalent to I 3. A. B. 1 C. 2 D. 3 E. 4 (xi) (3 points) A manufacturer s joint-cost function c for producing x units of product X and y units of product Y is given by c = 2x x + y + 6. What is the marginal cost with respect to the number of units of product X produced when x = 7 and y = 74? A. B. 35 6 C. 179 6 D. 1 12 E. 768 Remark: This is 17.2 #2 from Assignment 2. Page 4 of 14
(xii) (3 points) The value of 1 x x+y x 2 dz dy dx is A. B. 1 C. 3 1 D. 4 15 E. none of the above Remark: This is 17.9 #2 from Assignment 6. (xiii) (3 points) Exactly how many of the following mathematical statements are always true? The second order mixed-partial derivatives f xy and f yx of any function f : R 2 R are equal. If Q(x, y) is a polynomial function in the variables x and y (i.e. Q is nice ), then 1 x Q(x, y) dy dx = 1 y Q(x, y) dx dy. If a function f(x, y, z) has a relative maximum at (a, b, c) and (a, b, c) satisfy the constraint g(x, y, z) =, then there exists some λ R such that (a, b, c, λ ) is a critical point of F (x, y, z, λ) = f(x, y, z) λg(x, y, z). If the matrix A has more columns than rows, then the linear system defined by the matrix equation AX = B has infinitely many solutions. A. B. 1 C. 2 D. 3 E. 4 Page 5 of 14
Part II Full Solution Problems 2. (8 points) Let f(x, y) = x 2 +y 2 xy+x 3. Find the critical point(s) of f. For each critical point of f, use the second derivative test to classify it as a relative maximum, relative minimum, or a saddle point. Remark: This is 17.6 #14 from Assignment 6. Solution: We solve { fx (x, y) = 2x y + 3x 2 = f y (x, y) = 2y x = and find the critical points are (, ) and ( 1 2, 1 4 ). For the second derivative test, we find that f xx (x, y) = 2 + 6x f yy (x, y) = 2 f xy (x, y) = f yx (x, y) = 1. At (, ), f yy (, ) = 2 > D(, ) = 3 >, so (, ) is a relative minimum. At ( 1 2, 1 4 ), D( 1 2, 1 ) = 3 <, 4 so ( 1 2, 1 4 ) is a saddle point. Page 6 of 14
3. Let z = x 2 y 2. Sketch the level curves for Remark: This is 2.8 #3 from Assignment 5. (i) (2 points) z = Solution: The curve is x 2 y 2 = y = ±x. (ii) (2 points) z = 1 Solution: The curve is x 2 y 2 = 1. This is a hyperbola opening sideways. Page 7 of 14
Recall z = x 2 y 2 and you are asked to sketch the level curves for (iii) (2 points) z = 2 Solution: The curve is x 2 y 2 = 2. This is a hyperbola opening sideways. (iv) (2 points) z = 1 Solution: The curve is x 2 y 2 = 1 y 2 x 2 = 1. This is a hyperbola opening vertically. (v) (2 points) z = 2 Solution: The curve is x 2 y 2 = 2 y 2 x 2 = 2. This is a hyperbola opening vertically. Page 8 of 14
4. An oil company has two refineries and needs at least 8,, 14,, and 5, barrels of lowgrade, medium-grade, and high-grade oil respectively. Each day, Refinery I produces 2,, 3,, and 1, barrels of low-grade, medium-grade, and high-grade oil respectively, whereas Refinery II produces 1,, 2,, and 1, barrels of low-grade, medium-grade, and highgrade oil respectively. It costs $25, per day to operate Refinery I and $2, per day to operate Refinery II. Remark: This is 7.2 #18 from Assignment 2. (i) (7 points) How many days should each refinery be operated to satisfy the production requirements at minimum cost? (Assume that a minimum cost exists.) Solution: Let x and y be the number of days Refinery I and Refinery II are operated respectively. We minimize the cost function C = 25x + 2y subject to 2x + 1y 8 3x + 2y 14 1x + 1y 5 x, y. (constraint for low-grade oil) (constraint for medium-grade oil) (constraint for high-grade oil) The feasible region is unbounded and has four corner points: (, 8), (2, 4), (4, 1), and (5, ). Since we assumed a minimum exists and therefore know the minimum is attained at a corner point, we evaluate C at each corner point and find the minimum occurs at (4, 1). Therefore the company should operate Refinery I for 4 days and refinery II for 1 day for minimum cost. The feasible region (the unshaded part) and its four corner points. (ii) (1 point) What is the minimum cost? Solution: C(4, 1) = 25(4) + 2(1) = 12. Therefore, the minimum cost is $12,. Page 9 of 14
5. Let I = 5x dx dy where R = { (x, y) R 2 : y 2 x 3y, y 3 }. R Remark: This is 17.9 #16 from Assignment 6. (i) (3 points) Accurately sketch the region R over which the integration in I takes place. Solution: R is the unshaded part in the picture below: (ii) (4 points) Find the value of I. Solution: R 5x dx dy = = 3 3y 3 3 y 2 [ 5x 2 2 5x dx dy ] 3y y 2 dy 45y 2 = 5y4 2 2 dy [ ] 15y 3 3 = y5 2 2 = 45 2 243 2 = 81 Page 1 of 14
6. A company has taxable income $312,. The federal tax is 25% of the portion that is left after the provincial tax has been paid. The provincial tax is 1% of the portion that is left after the federal tax has been paid. Remark: This is 6.4 #27 from Assignment 3. (i) (4 points) Let x represent the company s federal tax and y represent the company s provincial tax. Write the system of equations satisfied by x and y. Solution: The system is { x =.25(312 y) y =.1(312 x). which simplifies to { x +.25y = 78.1x + y = 312. (ii) (4 points) Solve the system above using Cramer s rule. Any other method will not earn you any points. Solution: The equivalent the matrix equation is [ ] [ ] [ ] 1.25 x 78 =..1 1 y 312 Thus, by Cramer s rule, 78.25 312 1 x = 1.25 = 72 1 78.975 = 72 y =.1 312.1 1 1.25.1 1 = 234.975 = 24. Page 11 of 14
7. Suppose a manufacturer s production function is given by 16q = 65 4(l 4) 2 2(k 5) 2 where q is the number of units produces, and l and k are units of labour and capital respectively. The cost to the manufacturer is $8 per unit of labour and $16 per unit of capital. The selling price of the product is $64 per unit. Remark: This is 17.7 #2 from Assignment 6. (i) (2 points) Express the profit P as a function of q, l, and k. Solution: P = total revenue total cost = 64q (8l + 16k) = 64q 8l 16k (ii) (9 points) Using the method of Lagrange multipliers, find the values of q, l, and k that maximizes the profit P subject to the production constraint (assume a maximum exists). Solution: We maximize P = P (q, l, k) = 64q 8l 16k subject to g(q, l, k) = 16q 65 + 4(l 4) 2 + 2(k 5) 2 =. We look for the critical points to F (q, l, k, λ) = 64q 8l 16k λ ( 16q 65 + 4(l 4) 2 + 2(k 5) 2) by solving which gives F q = 64 16λ = F l = 8 8λ(l 4) = F k = 16 4λ(k 5) = F λ = 16q + 65 4(l 4) 2 2(k 5) 2 = λ = 4 l = 15 4 Therefore, maximum profit occurs at (q, l, k) = ( 251 64, 15 4, 4). k = 4 q = 251 64. Page 12 of 14
8. A Cobb-Douglas production function is a production function of the form P = Mx r y s where M, r, s are nonzero constants and r + s = 1. Remark: This is 17.2 #6 from Assignment 5. (i) (3 points) Show that P x = rp x. Solution: Since LHS = RHS, P x = rp x. LHS = P x RHS = rp x = rmx r 1 y s = rmxr y s x = rmx r 1 y s (ii) (3 points) Show that P y = sp y. Solution: LHS = P y RHS = sp y = smx r y s 1 = smxr y s y = smx r y s 1 Since LHS = RHS, P y = sp y. (iii) (3 points) Show that x P x + y P y = P. Solution: From the results above, x P x + y P y = xrp x + y sp y = (r + s)p = P since r + s = 1. Page 13 of 14
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