UNIVERSITY OF KWA-ZULU NATAL EXAMINATIONS: June 006 Solutions Subject, course and code: Mathematics 34 MATH34P Multiple Choice Answers. B. B 3. E 4. E 5. C 6. A 7. A 8. C 9. A 0. D. C. A 3. D 4. E 5. B 6. E 7. D 8. B 9. D 0. A
MCQ Solutions - 0. When the price of a certain commodity is R0, the quantity supplied per time period will be 60 units. When the price drops to R5, the producers will be willing to supply only 5 units per time period. Assuming that the supply function Q S is linear, the supply function will be given by: a Q S = 9 p + 30 9 b Q S = 9p 30 c Q S = 9p + 50 d Q S = 9p + 30 e Q S = 9 p + 40 9 Answer: B Putting the given information as ordered pairs p, Q S, we have 0, 60 and 5, 5 as the two points the line must go through. Then 5 60 slope = 5 0 = 9 Using the point-slope form of the linear equation, Q S Q S = mp p, and any of the two given points yields the equation Q S = 9p 30. The demand and supply functions for a certain commodity are given by Q D = 5p/3 and Q S = 0 p, respectively. The quantity exchanged at equilibrium is a 3 b 4 c 7 d 33 e 33/ Answer: B At equilibrium, Q D = Q S, so 5p/3 = 0 p, which gives p = 3. Putting p = 3 into either Q D or Q S yields Q E = 4. 3. Let A = 4 0, B =. The product AB is 4 0 a undefined Answer: E b 0 0 c 0 0 0 0 0 0 d 8 3 0 0 e 4 4 8 9 6 Since A is a 3 matrix and B is a 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 = c ij, where the entry c ij is found by multiplying row i of matrix A by column j of matrix B. We get 4 0 4 0 = + 4 + 0 + 4 + 04 4 + 00 4 + 0 + 4 + 0 + giving the matrix in e. 4. The solution to the matrix equation Ax = b is given by x = A b. Let A = 0 b =. The solution x is given by / / a b c d 4 / / 4 5 3 and e Answer: E 3 A = 5 4 3, so x = 5 4 0 =
5. The figure shows the graphs of the lines 4x y = 4, y x = and x = 3. The region satisfying the linear inequalities y is a A b B c C d D e none of these 4x y 4 y x x 3 x 0 y 0 0 8 6 4 D A B 0 x 0 3 4 C Answer: C Use the origin, 0, 0, as test point. For the inequalities 4x y 4 and y x, 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 = 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 = 3x + 4y subject to x + y 0 x + y 8 x y x, y 0 0 5 0 5 y 4, 0/3,4/3 A sketch of the lines corresponding to the constraints is shown on the right. The maximum value of P is 0 x 0 4 6 8 0 a 7 b 6/3 c 40 d 5 e 7 Answer: A The corner points of the feasible region are 0, 0, 0, 8, 4, and 0/3, 4/3. The value of P at the corner points is 40, 7, 0 and 6/3, respectively. So the maximum value of P is 7.
7. The Simplex method is being used to solve a maximisation problem. The following tableau has been reached x 3 is a surplus variable, x 4 is an artificial variable and x 5 is a slack variable: c j 8 0 M 0 c B Basis x x x 3 x 4 x 5 Solution Ratio M x 4 0 4 0 x 5 4 0 0 9 z j M M M M 0 4M c j z j + M 8 + M M 0 0 Which one of the following statements is true? a x must enter the basis and x 4 must leave the basis. b x must enter the basis and x 5 must leave the basis. c x must enter the basis and x 4 must leave the basis. d x must enter the basis and x 5 must leave the basis. e none of the these Answer: A Since + M in column for x is the biggest positive value in the c j z j row, x must enter the basis next. Taking the ratio of entries in the Solution column to the corresponding entries in the x column, we find the smallest ratio occurs in the row for x 4, so x 4 must leave the basis. 8. Let y = fx = a d y x = = y. The difference quotient x + 3 x is: x + 6 x + 3 x + 3 x + x + 3 b x + 3 4 c x + 6 x + x + x + e None of these. x + x + 3 x + 3 x = x + 6x + 9 x x x 6x x 6 x 9 x + x + 3 x + 3 x + 3 x + x + 3 x + x + 3 x + 3 x = x x x 6 x + x + 3 x + 3 x 6 x x + 3 x + x + 3 Answer: C 9. If fx = x + x x + 3, then the derivative f x equals: a x + x x + 3 + x + x b x + x c x + + x d x e x x + 3 + x + x Apply the product rule. Answer: A
0. If fx = x + 5x 7, then the derivative f x equals: a 8 x + 5x 8 4x + 5x b 7x + 5x 6 4x + 5 c 7x + 5x 6 d 7x + 5x 6 4x + 5 Apply the chain rule. Answer: D e none of these.. If ft = t, then the equation of the tangent line at, is: + 4 a y = 5 t + b y = 5 t + c y = 5 t + 7 5 d y = 5 t + 4 5 e none of these. f t = t f = = t +4 +4 5 y = 5 x y = 5 x + 7 5 Answer: C. Let the demand curve of a commodity be given by p = pq D. The price elasticity of demand is: a pq D p Q D Q D d Q Dp Q D pq D b pq D p c Q D p p Q D Q D Q D Q D e none of these. Price elasticity of demand is percentage change in Q D over the percentage change in p. Answer: A 3. Which statement applies to the function y = 3 x3 5x when x = 5? a y is at a maximum and concave down b y = 0 c y is at a point of inflection d y is at a minimum and concave up e None of these y x = x 0x = x x 5 y = 4x 0 When x = 5, y = 0 and y = 0, so y is at a minimum, and concave up. Answer: D 4. If z = 4x y + xe y, what is z x? a 8xy + xe y b 8x + c 8xy + e y d 8y + xe y e None of these z x = x Answer: E z = x x 8xy + ey = 8y
Section C - Financial Mathematics You are given the formulae: s n i = + in i and a n i = + i n. i Where necessary, round off to the nearest Rand. 5. What is the effective rate of interest of % p.a. compounded quarterly? a 5.8% b.46% c.5% d.8% e.57% i = + 0. 4 4 = 0. 4 6 Answer: B 6. 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 b R53 c R63 d R50 S = P + i n S = 500 + 0.06 48 = 635. 4 Answer: E e R635 7. Easy-Loans offers you a loan of R4 000. After 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 6.5% b.% c.50% d 6.07% e None of these S = P + i n 4500 = 4000 + x x = 6.07% Answer: D 8. A student buys a car for R 800. She makes a deposit of R 000 and agrees to pay the balance with equal monthly payments over years at an interest rate of 5% p.a. compounded monthly. How much are the monthly payments? a R60. 63 b R57. 4 c R44. 64 d R5863.40 e R49. 67 Use the present value formula: A = Ra n i i R = A + i n = 800 Answer: B 0.5 + 0.5 4 = 57. 4
9. A small business will yield R 00 000 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 R 70 66 b R3 8 54 c R500 000 d R 68 94 e None of these Quarterly Income = Quarterly Payment to sinking fund + Quarterly return on investment + 0.08 00000 = x 500000 + 0.x, Solution is:. 68 9 0 6. Answer: D 4 4 0.08 4 0. 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 R0 600 b R 667 c R0 55 d R 690 e R 577 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 4 payments is 900+ Present value of 3 payments = 900 + 900 + 0.05 3 0.05 = 0600. Answer: A
Long Questions -6 Solutions. Use the Gauss reduction technique to solve the following system of linear equations: x + x + x 3 = 3 x + x 3 = 0 x 4x 5x 3 = [5 Marks] Solution: We have 3 0 0 4 5 R R = R R R 3 = R 3 3 0 6 0 5 6 R / = R 3 0 / 3 0 5 6 R R = R R 3 5R = R 3 0 / 0 0 / 3 0 0 7/ 4 R 3 /7 = R 3 0 / 0 0 / 3 0 0 4 R R 3 / = R R R 3 / = R 0 0 0 0 5 0 0 4 So x =, x = 5 and x 3 = 4.. Use the Simplex algorithm to solve the following LP: maximise P = 3x + x subject to x + x 0 x + x x, x 0 Solution: [5 Marks] After adding slack variables x 3 in constraint and x 4 in constraint we get the following tableau: Tableau cj 3 0 0 c B Basis x x x 3 x 4 Solution Ratio 0 x 3 0 0 0 0 x 4 0 z j 0 0 0 0 0 c j z j 3 0 0 x enters basis and x 3 leaves basis. Tableau cj 3 0 0 c B Basis x x x 3 x 4 Solution Ratio 3 x 0 0 0 x 4 0 z j 3 3 3 0 30 c j z j 0 3 0 Above tableau is optimal. So solution is x = 0, x = 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 0 cents and 30 cents for each unit of product A and B respectively. There is R700 available to cover production costs and R0 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.x+0.3y = 0.8x+0.7y The required LP is then maximise P = 0.8x + 0.7y subject to 6x + 3y 700 [Production costs] 0.x + 0.3y 0 [Transportation costs] x, y 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 8% p.a. compounded monthly? Period Outstanding Balance Payment Interest Paid Principle Paid [6 Marks] + i n Use A = R i R = 59000 0.05.05 7 with A = 59 000, i = 0.8 = 345. 66 =.0 5, n = 6 = 7 months Period Outstanding Balance Payment Interest Paid Principle Paid 59 000 345. 66 885 460.66 58539.34 345. 66 878.09 467.57
5. The demand function for electric toothbrushes is p = 00 4q. The cost, in Rands, of producing q electric toothbrushes is C = 50 + 0q 0q + 4 3 q3 i Find the total revenue function. ii Find the profit function. iii How many toothbrushes should be made to maximise profit? [ Mark] [ Mark] [5 Marks] 5i R = 00 4q q = 00q 4q [] 5ii P = R C = 00q 4q 50 + 0q 0q + 4 3 q3 = 4 3 q3 + 6q + 80q 50 [] 5iii P = 4q + 3q + 80 0 = q 8q 0 0 = q + q 0 q = 0 nd derivative test: P 0 = 80 + 3 = 48. This confirms that q = 0 gives a maximum. [5]
6. i 5 4 x dx = ] 5 [ [4x x = 4 5 ] 5 [ 4 ] = 6 = 4 [3 Marks] ii x x 3 dx Let u = x 3 du dx = 6x, dx = du 6x x x 3 dx = x u du 6x = 6 u du = 6 u + c = x 3 + c [3 Marks] 6 iii 3 log e x dx g f = fg f g 3 log e x dx = 3x log x 3xdx = 3x log x 3x + c x [3 Marks] iv 3 x x dx Let u = x, then x = u + When x = 3, u = When x =, u = 3 x x dx = u + u dx = + du = [u + log u e u + c] = + log e + c + log e + c = + log e [3 Marks]