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UNIVERSITY OF KWA-ZULU NATAL EXAMINATIONS: June 2006 Solutions Subject, course and code: Mathematics 134 (MATH134P1) Multiple Choice Answers 1. B 2. B 3. E 4. E 5. C 6. A 7. A 8. C 9. A 10. D 11. C 12. A 13. D 14. E 15. B 16. E 17. D 18. B 19. D 20. A MCQ Solutions 1 - 20 1. When the price of a certain commodity is R10, the quantity supplied per time period will be 60 units. When the price drops to R5, the producers will be willing to supply only 15 units per time period. Assuming that the supply function (QS ) is linear, the supply function will be given by: (a) QS = 1 p + 9 Answer: B Putting the given information as ordered pairs (p, QS ), we have (10, 60) and (5, 15) as the two points the line must go through. Then 15 − 60 slope = =9 5 − 10 Using the point-slope form of the linear equation, QS − QS1 = m(p − p1 ), and any of the two given points yields the equation QS = 9p − 30 2. The demand and supply functions for a certain commodity are given by QD = 5p/3−1 and QS = 10−2p, respectively. The quantity exchanged at equilibrium is (a) 3 Answer: B At equilibrium, QD = QS , so 5p/3 − 1 = 10 − 2p, which gives p = 3. Putting p = 3 into either QD or QS yields QE = 4. 2 1 −1 −1 2 . The product AB is 3. Let A = 4 0 , B = 4 0 −2 −1 2 2 −2 2 1 0 0 −8 3 1 0 8 (d) (e) −4 −4 (a) undeﬁned (b) (c) 0 1 0 10 0 0 1 9 1 −6 0 0 1 Answer: E Since A is a 3 × 2 matrix and B is a 2 × 3 matrix (no. of columns of A is the same as no. of rows of B the matrix product AB is deﬁned and is the 3 × 3 matrix C = (cij ), where the entry cij is found by multiplying row i of matrix A by column j of matrix B. We get 2 4 −1 1 0 2 (2)(−1) + (1)(4) (2)(−1) + (1)(0) (2)(2) + (1)(−2) (4)(−1) + (0)(0) (4)(2) + (0)(−2) = (4)(−1) + (0)(4) (−1)(−1) + (2)(4) (−1)(−1) + (2)(0) (−1)(2) + (2)(−2) (b) 4 (c) 27 (d) 33 (e) 33/11 130 9 (b) QS = 9p − 30 (c) QS = −9p + 150 (d) QS = 9p + 30 (e) QS = − 1 p + 9 140 9 −1 −1 2 4 0 −2 giving the matrix in (e). 4. The solution to the matrix equation Ax = b is given by x = A−1 b. Let A = b= (a) 0 −1 −2 −4 . The solution x is given by (b) 1/11 2/11 (c) −1 −2 (d) −1/11 −2/11 (e) 1 2 4 −2 5 −3 and Answer: E 1 A−1 = − 2 −3 −5 2 4 , so x = − 1 2 −3 −5 2 4 0 −1 = 1 2 5. The ﬁgure shows the graphs of the lines 4x − y = 4, y − x = 1 and x = 3. The region satisfying the linear inequalities y 4x − y y − x x x y is (a) A (d) D (b) B (c) C (e) none of these ≥ ≥ ≤ ≥ ≥ 4 1 3 0 0 10 8 6 D 4 2 0 0 1 2 3 4 C A B x Answer: C Use the origin, (0, 0), as test point. For the inequalities 4x − y ≥ 4 and y − x ≥ 1, (0, 0) is not feasible. So feasible region is on the side of the lines that does not contain (0, 0), i.e., feasible region is above the line y − x = 1 (eliminating regions A and B) and to the right of the line 4x − y = 4 (eliminating region D). Combining this with the constraint x ≤ 3 and the non-negativity constraints gives C as the feasible region. 6. Consider the following linear programming problem: maximise P subject to = 3x 2x 2x x + 4y + y + y − y x, y ≥ 10 ≤ 18 ≤ 2 ≥ 0 20 15 10 5 0 0 2 4 6 8 (4,2) (20/3,14/3) x 10 y A sketch of the lines corresponding to the constraints is shown on the right. The maximum value of P is (a) 72 Answer: A (b) 116/3 (c) 40 (d) 15 (e) 27 The corner points of the feasible region are (0, 10), (0, 18), (4, 2) and (20/3, 14/3). The value of P at the corner points is 40, 72, 20 and 116/3, respectively. So the maximum value of P is 72. 7. The Simplex method is being used to solve a maximisation problem. The following tableau has been reached (x3 is a surplus variable, x4 is an artiﬁcial variable and x5 is a slack variable): cB −M 0 cj Basis x4 x5 zj cj − zj 1 x1 2 1 −2M 1 + 2M 8 x2 1 4 −M 8+M 0 x3 −1 0 M −M −M x4 1 0 −M 0 0 x5 0 1 0 0 Solution 4 9 −4M Ratio Which one of the following statements is true? (a) x1 must enter the basis and x4 must leave the basis. (b) x1 must enter the basis and x5 must leave the basis. (c) x2 must enter the basis and x4 must leave the basis. (d) x2 must enter the basis and x5 must leave the basis. (e) none of the these Answer: A Since 1 + 2M (in column for x1 ) is the biggest positive value in the cj − zj row, x1 must enter the basis next. Taking the ratio of entries in the “Solution” column to the corresponding entries in the x1 column, we ﬁnd the smallest ratio occurs in the row for x4 , so x4 must leave the basis. 8. Let y = f (x) = 1 ∆y . The diﬀerence quotient is: 2 (x + 3) ∆x (b) − 2(x + 3) (x + 3)4 (c) −2x − 6 − ∆x (x + 3)2 (x + ∆x + 3)2 2x + 6 (x + 3)2 (x + ∆x + 3)2 2x + 6 (d) (x + 1)2 (x + ∆x + 2)2 (a) (e) None of these. ∆y = ∆x 1 1 − (x + ∆x + 3)2 (x + 3)2 1 = ∆x (x + 3)2 − (x + ∆x + 3)2 (x + ∆x + 3)2 (x + 3) 1 = ∆x 2 1 ∆x = x2 + 6x + 9 − x2 − 2x∆x − 6x − ∆x2 − 6∆x − 9 (x + ∆x + 3)2 (x + 3) 2 −2x − ∆x − 6 (x + ∆x + 3)2 (x + 3) 2 Answer: C 9. If f (x) = (x + 2)2 (x2 − 2x + 3), then the derivative f (x) equals: (a) 2(x + 2)(x2 − 2x + 3) + (x + 2)2 (2x − 2) (c) 2(x + 2) + (2x − 2) (e) 2(x2 − 2x + 3) + (x + 2)2 (2x − 2) Apply the product rule. Answer: A (b) 2(x + 2)(2x − 2) (d) 2(2x − 2) 10. If f (x) = (2x2 + 5x)17 , then the derivative f (x) equals: (a) 1 2 18 (2x + 5x)18 (4x + 5x) (b) 17(2x2 + 5x)16 (4x) + 5 (e) none of these. (c) 17(2x2 + 5x)16 (d) 17(2x2 + 5x)16 (4x + 5) Apply the chain rule. Answer: D 11. If f (t) = t2 1 , then the equation of the tangent line at (1, 1) is: +4 1 (b) y = 5 t + 1 2 (c) y = − 25 t + 27 25 2 (a) y = − 25 t + 1 1 (d) y = 5 t + 4 5 (e) none of these. −2 (1+4)2 2 = − 25 f (t) = −2t (t2 +4)2 f (1) = 2 y − 1 = − 25 (x − 1) 2 y = − 25 x + 27 5 Answer: C 12. Let the demand curve of a commodity be given by p = p(QD ). The price elasticity of demand is: (a) − (d) − p(QD ) p (QD )QD QD p (QD ) p(QD ) (b) p(QD ) p (QD )QD (c) QD (p) p (QD ) QD (e) none of these. Price elasticity of demand is percentage change in QD over the percentage change in p. Answer: A 13. Which statement applies to the function y = 2 x3 − 5x2 when x = 5? 3 (a) y is at a maximum and concave down (d) y is at a minimum and concave up (b) y = 0 (e) None of these (c) y is at a point of inﬂection y (x) = 2x2 − 10x = 2x (x − 5) y = 4x − 10 When x = 5, y = 0 and y = 10, so y is at a minimum, and concave up. Answer: D 14. If z = 4x2 y + xey , what is (a) 8xy + xey (d) 8y + xey ∂2z ∂ ∂z = 2 ∂x ∂x ∂x Answer: E ∂2z ? ∂x2 (b) 8x + 1 (e) None of these (c) (8xy + ey ) 2 = ∂ (8xy + ey ) = 8y ∂x Section C - Financial Mathematics You are given the formulae: sn i = (1 + i) − 1 i n and an i = 1 − (1 + i) i −n . Where necessary, round oﬀ to the nearest Rand. 15. What is the eﬀective rate of interest of 11% p.a. compounded quarterly? (a) 15.18% i= 1+ 0.11 4 4 (b) 11.46% − 1 = 0. 114 62 (c) 21.15% (d) 11.28% (e) 11.57% Answer: B 16. Paul invests R500 in an account earning 6% p.a. compounded monthly. To what amount would it accumulate at the end of 4 years, rounded to the nearest Rand? (a) R598 (d) R510 S = P (1 + i) S = 500 1 + Answer: E 17. Easy-Loans oﬀers you a loan of R4 000. After 2 years you must pay Easy-Loans R4 500. What is the nominal interest rate that Easy-Loans are charging, if interest is compounded annually? (a) 6.25% (d) 6.07% S = P (1 + i) x = 6.07% Answer: D n 2 n 0.06 48 12 (b) R531 (e) R635 (c) R631 = 635. 24 (b) 11.11% (e) None of these (c) 12.50% 4500 = 4000 (1 + x) 18. A student buys a car for R12 800. She makes a deposit of R1 000 and agrees to pay the balance with equal monthly payments over 2 years at an interest rate of 15% p.a. compounded monthly. How much are the monthly payments? (a) R620. 63 (d) R5863.40 (b) R572. 14 (e) R491. 67 i 0.15 12 −24 + 0.15 12 (c) R424. 64 Use the present value formula: A = Ran R=A i 1 − (1 + i) −n = 11800 1− 1 = 572. 14 Answer: B 19. A small business will yield R 200 000 at the end of each quarter for 6 years after which it will can be sold for R500 000. What should an investor pay for the business if a 40% p.a. return on capital is required, and the sinking fund, into which quarterly payments are made, earns 8% p.a. compounded quarterly? (a) R1 170 266 (d) R1 628 914 (b) R3 812 524 (e) None of these (c) R500 000 Quarterly Income = Quarterly Payment to sinking fund + Quarterly return on investment 200000 = (x − 500000) Answer: D 20. Rent of R900 per month is payable at the beginning of each month. If the interest rate is 5% p.a. compounded monthly, what is the cash equivalent of two years’ of rent paid in advance? (a) R20 600 (d) R21 690 (b) R22 667 (e) R22 577 (c) R20 515 (1+ 0.08 ) 4 0.08 4 24 −1 −1 + 0.1x, Solution is: 1. 628 9 × 106 . This is an annuity due and we are asked for the present value of the payments. The ﬁrst payment is already a present value, so the cash equivalent of 24 payments is 900+ Present value of 23 payments = 900 + 900 Answer: A 1−(1+ 0.05 ) 12 0.05 12 −23 = 20600. Long Questions 1 -6 Solutions 1. Use the Gauss reduction technique to solve the following system of linear equations: x1 2x1 x1 + x2 + − 4x2 + x3 x3 − 5x3 = 3 = 0 = 2 [5 Marks] Solution: We have 1 1 1 3 1 1 1 3 2 0 1 0 −→ 2R1 − R2 = R2 0 2 1 6 R1 − R3 = R 3 1 −4 −5 2 0 5 6 1 −→ R2 /2 = R2 1 0 0 −→ 2R3 /7 = R3 1 1 5 R1 − R 2 = R 1 1 0 1/2 1 3 0 0 1 1/2 3 1/2 3 −→ R3 − 5R2 = R3 6 1 0 0 7/2 −14 1 0 1/2 0 2 R1 − R3 /2 = R1 1 0 0 0 1 1/2 3 −→ R2 − R3 /2 = R2 0 1 0 5 0 0 1 −4 0 0 1 −4 So x1 = 2, x2 = 5 and x3 = −4. 2. Use the Simplex algorithm to solve the following LP: maximise subject to P = 3x1 x1 x1 + 2x2 + x2 ≤ 10 + 2x2 ≤ 12 x1 , x2 ≥ 0 [5 Marks] Solution: After adding slack variables x3 (in constraint 1) and x4 (in constraint 2) we get the following tableau: Tableau 1 cj cB Basis 0 x3 0 x4 zj cj − zj 3 x1 1 1 0 3 ↑ 2 x2 1 2 0 2 0 x3 1 0 0 0 0 x4 0 1 0 0 Solution 10 12 0 Ratio 10 12 ← x1 enters basis and x3 leaves basis. Tableau 2 cj cB Basis 3 x1 0 x4 zj cj − zj 3 x1 1 0 3 0 2 x2 1 1 3 −1 0 x3 1 −1 3 −3 0 x4 0 1 0 0 Solution 10 2 30 Ratio Above tableau is optimal. So solution is x1 = 10, x2 = 0 maximises P to 30. 3. A small manufacturer produces two products, A and B. The production costs per unit of A and B are R6 and R3, respectively, and the corresponding selling prices are R7 and R4. In addition, there are transportation costs of 20 cents and 30 cents for each unit of product A and B respectively. There is R2700 available to cover production costs and R120 to cover transportation costs. Set up an LP to help the manufacturer decide how many units of each product must be produced in order to maximise proﬁt. [Do Not Solve The LP - no extra marks will be given for solving the LP.] [5 Marks] Solution: Let x = number of units of product A produced y = number of units of product B produced Then proﬁt (total revenue−total costs) is given by P = (7x+4y)−(6x+3y)−(0.2x+0.3y) = 0.8x+0.7y The required LP is then maximise subject to P = 0.8x + 0.7y 6x + 3y 0.2x + 0.3y x, y ≤ 2700 [Production costs] ≤ 120 [Transportation costs] ≥ 0 4. Complete the ﬁrst two lines of an amortization schedule for a debt of R59 000 being paid oﬀ over 6 years by monthly paymens and with an interest rate of 18% p.a. compounded monthly? Period 1 2 Outstanding Balance Payment Interest Paid Principle Paid [6 Marks] Use A = R 1 − (1 + i) i 0.015 1−(1.015)−72 −n with A = 59 000, i = 0.18 12 = .0 15, n = 6 (12) = 72 months R = 59000 = 1345. 66 Period 1 2 Outstanding Balance 59 000 58539.34 Payment 1345. 66 1345. 66 Interest Paid 885 878.09 Principle Paid 460.66 467.57 5. The demand function for electric toothbrushes is p = 100 − 4q. The cost, in Rands, of producing q electric toothbrushes is 4 C = 50 + 20q − 20q 2 + q 3 3 (i) Find the total revenue function. (ii) Find the proﬁt function. (iii) How many toothbrushes should be made to maximise proﬁt? [1 Mark] [1 Mark] [5 Marks] 5(i) R = (100 − 4q) q = 100q − 4q 2 [1] 5(ii) P = R − C 4 = 100q − 4q 2 − 50 + 20q − 20q 2 + 3 q 3 = − 4 q 3 + 16q 2 + 80q − 50 3 [1] 5(iii) P = −4q 2 + 32q + 80 0 = q 2 − 8q − 20 0 = (q + 2) (q − 10) q = 10 2nd derivative test: P (10) = −8(10) + 32 = −48. This conﬁrms that q = 10 gives a maximum. [5] 5 6. (i) 1 (4 − x) dx = 4x − x2 2 5 = 4 (5) − 1 52 2 − 4 (1) − 12 2 = 16 − 12 = 4 [3 Marks] (ii) 2 dx (2x3 − 1) Let u = 2x3 − 1 du du = 6x2 , dx = 2 dx 6x x2 x2 du = 2 dx = u2 6x2 (2x3 − 1) x2 1 −2 1 1 u du = − u−1 + c = − 2x3 − 1 6 6 6 −1 +c [3 Marks] (iii) 3 loge x dx g f = fg − fg 1 3xdx = 3x log x − 3x + c x [3 Marks] 3 loge x dx = 3x log x − 3 (iv) x dx (x − 1) 2 Let u = x − 1, then x = u + 1 When x = 3, u = 2 When x = 2, u = 1 3 2 2 x u+1 1 dx = dx = 1+ (x − 1) (u) u 2 1 1 = 2 + loge 2 + c − (1 + loge 1 + c) = 1 + loge 2 du = [u + loge u + c]1 [3 Marks] 2