Student Number: SOLUTION Page 1 of 14

Transcription

1 Student Number: SOLUTION Page 1 of 14 QUEEN S UNIVERSITY FACULTY OF ARTS AND SCIENCE DEPARTMENT OF MATHEMATICS AND STATISTICS MATH126 December Examination December 14th, 2009 Instructors: P. Li (A), A. Momeni (B) Instructions: This is a 3-hour exam. There are 5 sections worth a total of 100 marks as indicated in the box below. Answer all questions in the space provided. If you need more room, answer on the back of the previous page. Show all your work and explain how you arrived at your answers, unless explicitly told to do otherwise. Except where a decimal answer is asked for, it is preferable to leave answers in the form π, e 2, etc. However, do any obvious simplification. For example, = 25 6 or 17 (x + 1)2, and 6 x + 1 are permitted. = x+1. Only CASIO FX-991 or Gold/Bule Sticker calculators Please Note: Proctors are unable to respond to queries about the interpretation of exam questions. Do your best to answer exam questions as written. Question Possible Received Sections I 45 Section II 29 Section III 8 Section VI 10 Section V 8 Total 100

2 Student Number: SOLUTION Page 2 of 14 Section I: Short Answer Questions (15 questions, 3 marks each) Solve the following questions and write your answer in the box provided in each question. Full marks will be given for correct answer placed in the box. You do not need to simplify your answer. (1) Evaluate lim x 2 x 2 + x 6 x 2 Since x 2 + x 6 = (x 2)(x + 3) x 2 + x 6 lim x 2 x 2 = lim x 2 x + 3 = 5 5 (2) Evaluate lim x 1 x x 2 2x 2 7. Using L Hôpital s rule, we have 1 x x 2 lim x 2x 2 7 = lim x 1 2x 4x 2 = lim x 4 = (3) Find the derivative of y = arctan y = 1 ( ) 2 ( 1x ) x ( ) 1. x = 1 x x 2 + 1

3 Student Number: SOLUTION Page 3 of 14 Section I (continued) (4) Find the equation of the tangent line to the curve y = x ln x at the point (1, 1). Write your answer in slope-intercept form, i.e, y = mx + b. y = 3x 2 + 4/x. At x = 1, y = 7 Then the equation is y = 7x 6 y = 7(x 1) + 1, i.e., y = 7x 6. (5) Find the derivative of y = y = x cos x sin x x 2 sin x x e x2 2x ex2 x cos x sin x x 2 2xe x2 (6) Use implicit differentiate to find dy dx. y = x 2 ln y + 2e x Differentiating both sides, we have dy dx = 2x ln y + 1 x2 y dy dx + 2ex Solving dy, we obtain the answer. dx 2x ln y + 2e x 1 x2 y

4 Student Number: SOLUTION Page 4 of 14 Section I (continued) (7) If an investor has a choice of investing money of 6% compounded daily or 6.125% compounded monthly, which is a better choice? The effective rates are: At the rate 6%, r e1 = ( /365) = At the rate 6.125%, r e2 = ( /12) 12 1 = r e2 > r e % compounded monthly (8) At what nominal rate of interest, compounded yearly, will money double in 10 years? (1 + r) 10 = 2 r = = % (9) Use L Hôpital s rule to evaluate the following limit x + cos(2x) e x lim x 0 x x + cos(2x) e x lim x 0 x 1 2 sin(2x) e x = lim x 0 1 = 0 0

5 Student Number: SOLUTION Page 5 of 14 Section I (continued) (10) What is the present value of $1000 due after three years if the interest rate is 9% compounded monthly? 1000( /12) 36 = (11) For an interest rate of 4% compounded monthly, find an expression of the present value of an annuity of $150 at the end of each month for eight months and $175 thereafter at the end of each month for a further two years. the periodic rate = 0.04/12 = In the first 8 months, the periodic payment is 150, the number of periods is 8, so the 175a a present value is 150a In the next two years, the periodic payment is 175, the number of periods is 24, so the present value is 175a a Hence, the present value of the annuity is 150a a a (12) If a manufacturer s average cost is c = 0.01q q what is the marginal cost when 50 units are produced. The cost function is c = cq = 0.01q 2 + 5q The marginal cost is c = 0.02q + 5 at q = 50, c = 6 6

6 Student Number: SOLUTION Page 6 of 14 Section I (continued) (13) If a demand equation for a manufacturer s product is p = 1000 q + 5 find the marginal revenue when q = 45. The revenue is r = pq = 1000q/(q + 5) The marginal revenue is: r = 5000/(q + 5) 2 at q = 45, r = 2 2 (14) Determine the value(s) of x at which the function f = 1 3 x x2 2x + 1 attains a maximum value over the interval [ 1, 2]. f = x 2 + x 2 = (x + 2)(x 1) = 0 x = 1 critical values are x = 1 and x = 2 x = 2 is ignored because it is outside of the interval [ 1, 2] comparing values of the function: f( 1) = 19/6, f(1) = 1/6, f(2) = 5/3 The maximum is at x = 1 (15) Find all critical values of the function y = x 2 e x y = 2xe x + x 2 e x = xe x (2 + x) = 0 x = 0, x = 2 x = 0, x = 2

7 Student Number: SOLUTION Page 7 of 14 Section II: Graphical Problems (4 questions, 29 marks) 1. [5 marks] The graph of the function y = f(x) is given below. Find the following values and sketch the graph of the derivative of y = f(x) over the interval [ 2, 5]. f( 1) = 1 f( 1 + h) f( 1) lim h 0 h = 2 f(1 + h) f(1) lim h 0 h = f (1) = 0 f (4) = 0 f (4) = 0 y y = f(x) x y e e e x e

8 Student Number: SOLUTION Page 8 of 14 Section II: Graphical Problems (continued) 2. [6 marks] The following graph shows cost C(q) and revenue R(q). $ 400 C(q) Break-even v Break-even R(q) v q (thousands) (a) For what production level is the profit positive? Negative? Explain your answer. Mark the break-even point(s) on the graph. Solution: The profit is positive if 6 < q < 14 because the revenue is greater than the cost. The profit is negative if q < 6 or q > 14 because the revenue is smaller than the cost. The break-even points are at q = 6 and q = 14. (b) Find a construction on the above graph which maximize the profit. Estimate the production level at which the profit is maximized. Explain your answer. Solution: The profit attains maximum when q = 11 because MR = MC and P > 0. It can also be seen as follows: P = R C, so the profit is represented by the vertical distance between R and C. By the given graph, at q = 11, R > C and the vertical distance between R and C is maximum. Therefore P is maximum at q = 11.

9 Student Number: SOLUTION Page 9 of 14 Section II: Graphical Problems (continued) 3. [8 marks]the marginal revenue and marginal cost for a certain item are graphed below. (a) Analyze each of the following quantities and explain the economic significance of the quantities (in terms of profit, marginal revenue and marginal cost, etc.). (b) Find the quantity at which the profit attains maximum for the company? Explain your answer in detail. (c) Give a rough sketch of the profit graph P = P (q). Mark all significant points on your graph and explain the shape of your graph. You may continue on next page if you need more space. (A) q = q 1 (B) q = q 2 (C) q = q 3 $/unit MR MC q 1 q 2 q 3 q (units) Solution: (a) At q = q 1, MR = MC. The revenue of producing the next unit is equal to the cost of that unit. Since MR is increasing and MR > MC passing q 1, A is the point at which the company receives minimum profit. The profit will increase after q 1. At q = q 2, MR > MC and MR is peaked. Note that MC is constant. So the company expect maximum return from the next unit produced. After q 2, MR decreases, but still greater than M C. It implies a positive return from additional unit produced. so the profit keeps increasing. At q = q 3, MR = MC. Since MR deceases and MR < MC when q > q 3, the company is no longer making money from addition unit produced. The profit is maximized at q = q 3, and after q 3, the profit will decrease. (b) The profit is maximum at q 3. It can be seen as follows. The profit is P = R C. The critical points occur when P = 0, that is R C = MR MC = 0. According to the graph, the critical points are at q = q 1 and q = q 3. To determine which one is maximum, we use the second derivative test. Take P = MR MC. Note that MC = 0. It follows P = MR. By the graph, at q = q 1, P = MR > 0, so P attains a minimum at q = q 1. At q = q 3, P = MR < 0, so P attains a maximum at q = q 3.

10 Student Number: SOLUTION Page 10 of 14 Section II: Graphical Problems (continued) (c) As discussed in part (b), there are two critical points at q = q 1 and q = q 3. Since MR < MC when q < q 1 and q > q 3 and MR > MC if q 1 < q < q 3, we have the following chart q < q 1 q = q 1 q 1 < q < q 3 q = q 3 q > q 3 P = MR MC + y dec. relative min inc. relative max dec. The second derivative P = R C = MR MC = MR since MC = 0. Let P = 0 we have MR = 0. By the graph, MR = 0 if q = q 2. So q = q 2 is a possible inflection point. It is easy to know, from the graph, MR > 0 if q < q 2 and MR < 0 if q > q 2. Therefore the graph of P is concave up if q < q 2 and concave down if q > q 2. The point at q = q 2 is an inflection point. a sketch of the profit function is P r r r q 1 q 2 q 3 q

11 Student Number: SOLUTION Page 11 of 14 Section II: Graphical Problems (continued) 4. [10 marks] Sketch the graph of y = 2x 3 9x x. (a) Find all intercept points. Solution: x-intercepts: Let y = 0, that is, 2x 3 9x x = x(2x 2 9x + 12) = 0. It yields x = 0 and 2x 2 9x + 12 = 0. But the equation 2x 2 9x + 12 = 0 has no solution since the determinant b 2 4ac = ( 9) 2 4(2)(12) < 0. So The x-intercept is x = 0. y-intercept: Let x = 0. Then y = 0. (b) Find all critical values. Find all relative maxima and relative minima (if there are any). Find interval(s) on which the function is increasing or decreasing. Solution: Since y = 6x 2 18x + 12 = 6(x 2 3x + 2) = 6(x 1)(x 2), and critical values occur when y = 0. Hence x = 1 and x = 2 are critical values. Now we have the following table. x < 1 x = 1 1 < x < 2 x = 2 2 < x 6(x 1) + + x 2 + y + + y inc. relative max dec. relative min inc. There is one relative maximum at x = 1. There is one relative minimum at x = 2. The function increases over intervals (, 1) and (2, ). and decreases over the interval (1, 2). (c) Find interval(s) on which the function is concave up or concave down. Solution: y = 12x 18 = 12(x 3/2) = 0. We have x = 3/2. x < 3/2 x = 3/2 x > 3/2 y + y concave inflection concave down point up (d) Sketch the graph. y r r r x

12 Student Number: SOLUTION Page 12 of 14 Section III: Equilibrium (8 marks) Supply and demand equations for certain product are and 3q 200p = 0 3q + 100p 1800 = 0 respectively, where p represents the price per unit in dollars and q represents the number of units sold per time period. Find the equilibrium price when a tax of 27 cents per unit is imposed on the supplier. Solution: The supply equation is p = q + 9 and the demand equation is p = 3 q If $0.27 is applied to the supply, then the new supply equation is p = 3 q To find the equilibrium point, we solve the equation 3 3 q = q The solution is q = 194. Hence the equilibrium price is p =

13 Student Number: SOLUTION Page 13 of 14 Section VI: Related Rates (10 marks) The demand equation for tuna is pq 3/2 = where q is the number of pounds of tuna that can be sold in one month at the price of p dollars per pound. Currently the demand of tuna is 900 pounds per month and is increasing at a rate of 100 pounds per month each month. (a) How fast is the price changing? Solution: The values we know are the demand: q = 900 the rate of change of demand with respect to the time t: dq dt = 100 We want to find dp dt. Differentiating pq 3/2 = with respect to t, we have Solving for dp dt yields dp dt q3/2 + p 3 dq q1/2 2 dt = 0 dp dt = 1 3pq 1/2 dq q 3/2 2 dt = 3p 2q dq dt At q = 900, p = Then dp dt 0.31 So the price is dropping at the rate of 31 cents per pound each month. (b) At what rate is the revenue changing? Will the revenue increase or decrease? Solution: The revenue is r = pq. Differentiating both sides with respect to the time t and plugging into all values, we have dr dt = dp q + pdq dt dt The revenue is decreasing. 50 = =

14 Student Number: SOLUTION Page 14 of 14 Section V: Optimization Problem (8 marks) A manufacturer can produce at most 120 units of a certain product each year. The demand equation for the product is p = q 2 100q and the manufacturer s average cost function is c = 2 3 q2 40q q (a) Find the revenue function R and the cost function C. What is the profit function P? Solution: R = pq = q 3 100q q C = cq = 2 3 q3 40q P = R C = 1 3 q3 60q q for q 120. (b) Determine the production level at which the profit is a maximum. Solution: P = q 2 120q = (q 40)(q 80) = 0, we have q = 40 and q = 80. Since P (40) = , P (80) = and P (120) = 86000, the profit is maximum when q = 120.