FACULTY OF ARTS AND SCIENCE University of Toronto FINAL EXAMINATIONS, APRIL 2012 MAT 133Y1Y Calculus and Linear Algebra for Commerce

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1 FACULTY OF ARTS AND SCIENCE University of Toronto FINAL EXAMINATIONS, APRIL 0 MAT 33YY Calculus and Linear Algebra for Commerce Duration: Examiners: 3 hours N. Francetic A. Igelfeld P. Kergin J. Tate LEAVE BLANK FAMILY NAME: GIVEN NAME: STUDENT NO: SIGNATURE: Question Mark MC /45 B /3 B / B3 /0 B4 /0 B5 /0 TOTAL NOTE:. Aids Allowed: A non-graphing calculator, with empty memory, to be supplied by student.. Instructions: Fill in the information on this page, and make sure your test booklet contains 4 pages. 3. This exam consists of 5 multiple choice questions, and 5 written-answer questions. For the multiple choice questions you can do your rough work in the test booklet, but you must record your answer by circling the appropriate letter on the front page with your pencil. Each correct answer is worth 3 marks; a question left blank, or an incorrect answer or two answers for the same question is worth 0. For the writtenanswer questions, present your solutions in the space provided. The value of each written-answer question is indicated beside it. 4. Put your name and student number on each page of this examination. ANSWER BOX FOR PART A Circle the correct answer. A. B. C. D. E.. A. B. C. D. E. 3. A. B. C. D. E. 4. A. B. C. D. E. 5. A. B. C. D. E. 6. A. B. C. D. E. 7. A. B. C. D. E. 8. A. B. C. D. E. 9. A. B. C. D. E. 0. A. B. C. D. E.. A. B. C. D. E.. A. B. C. D. E. 3. A. B. C. D. E. 4. A. B. C. D. E. 5. A. B. C. D. E. Page of 4

2 PART A. MULTIPLE CHOICE. [3 marks] If P Q = R where P = and Q = then r 4 is A. 4 B. 5 C. 6 D. 7 E. undefined because P and Q cannot be multiplied together. [3 marks] x + e k if x If f(x) = x x if x > x is continuous at x = then A. k = B. k = 0 C. k = D. k = E. there is no real number k such that f is continuous at x = Page of 4

3 3. [3 marks] If y = x ln x, then y (e) = A. B. e e C. e D. E. e 4. [3 marks] The function f(x) = xe x on the inverval [, ] A. has an absolute minimum at x = and an absolute maximum at x = B. has an absolute minimum at x = and an absolute maximum at x = C. has an absolute minimum at x = and an absolute maximum at x = D. has an absolute maximum at x = and no absolute minimum E. has an absolute minimum at x = and no absolute maximum Page 3 of 4

4 5. [3 marks] If the demand equation is p = q3 50q q, for what values of q is the marginal revenue ( dr dq ) decreasing? A. (0, 00) B. (00, 00) C. (00, ) D. All values E. None 6. [3 marks] e x x x lim x 0 x 3 A. undefined B. 0 C. D. 6 E. 4 3 is Page 4 of 4

5 7. [3 marks] If Newton s method is used to estimate a solution to the equation beginning with x =, then x is closest to A B..33 C..39 D..4 E..43 (x ) = ln x 8. [3 marks] e ln x x dx = A. B. C. D. e E. e Page 5 of 4

6 9. [3 marks] x + x dx = A. (x + ) ln x + C B. 5 ln x + C C. [ln x + ln x + ln x ] + C D. [ln x ln x + ] + C E. [3 ln x ln x + ] + C 0. [3 marks] A town s population is expected to grow exponentially so that it will double every 50 years. population today is 0,000, then in 0 years its population will be A., 039 B., 755 C. 0, 8 D., 487 E. 0, 43 If its Page 6 of 4

7 . [3 marks] If z = (x + y ) ln(3x + y), then when x = 0 and y = e, A. 0 B. C. e D. e E. 3e z x =. [3 marks] If y + z = xyz +, then when x = y = z =, A. B. 0 C. D. E. z y = Page 7 of 4

8 3. [3 marks] If x = 3r + 5s, y = r + 3s, and z = x + y, then when r = s =, A. 64 B. 0 C. 4 D. 68 E. 39 z r = 4. [3 marks] The quantities q A and q B of goods A and B, which are sold when the respective unit prices of the goods are p A and p B, are q A (p A, p B ) = 00 5p A + 0p B p B q B (p A, p B ) = 00 6p A 3p B + p A The two goods are complementary A. when p A > 3 and p B < 5 B. when p A > 3 and p B > 5 C. when p A < 3 and p B > 5 D. for no values of p A and p B E. when p A < 3 and p B < 5 Page 8 of 4

9 5. [3 marks] The critical point of is A. (, ), which is a saddle point B. (, ), which is a local max C. (, ), which is a local min f(x, y) = x + xy y x 6y + 3 D. (, ), but the second derivative test cannot determine its nature E. not defined, because the partial derivatives are never both zero Page 9 of 4

10 PART B. WRITTEN-ANSWER QUESTIONS B. [3 marks] At the end of December, 000, Henry invested $00,000 in an annuity which could be refunded at any time and which made 0 equal semiannual payments at the ends of June and December from 00 to 00 inclusive at 6% compounded semi-annually. (a) [4 marks] What was the amount of each payment Henry received from the annuity? (b) [4 marks] At the end of 007, just after the annuity s 4 th payment, Henry refunded it, that is, sold it back to the bank for its current value (value at the end of 007). How much money did he receive? (c) [5 marks] At the beginning of 008, Henry bought as many Acme bonds as he could with the money he received from the sale of the annuity. An Acme bond had a face value of $000, 6 semiannual coupons worth $30 each until maturity at the end of 00, and an annual yield rate of 7%. How many Acme bonds could Henry buy? Page 0 of 4

11 B. [ marks] Given: At the present time, 800 people pay $9 each to take a commuter train to work. The number of people willing to ride the train at a price p is given by: (a) [4 marks] At present, is the demand elastic or inelastic? q = 600(6 p) (b) [ marks] By examining dr dp revenue (r). at the present time, determine whether an increase in price will increase or decrease (c) [5 marks] Find the price that should be charged in order to maximize revenue. (d) [ mark] At the price found in c), will an increase in price result in an increase or decrease in revenue? Why? Page of 4

12 B3. [0 marks] (a) [4 marks] Sketch the functions y = x and y = x on the axes below, labelling the points of intersection on the graph (b) [6 marks] Shade the region(s) between the graphs of the two functions from x = to x = 5 and evaluate the area of the shaded figure. [Express your final answer as a single number.] Page of 4

13 B4. [0 marks] Find y(x) explicitly in terms of x such that dy dx = xey and y(0) =. [You may assume that x is always between e and e.] Page 3 of 4

14 B5. [0 marks] Use the method of Lagrange multipliers to find all critical points of the problem maximize 4x + y subject to the contraint x + xy = 7 [No marks will be given for any other method.] Make sure to specify the values of the Lagrange multiplier. Page 4 of 4

15 PART A. MULTIPLE CHOICE. [3 marks] If P Q = R where P = and Q = then r 4 is A. 4 B. 5 C. 6 D. 7 E. undefined because P and Q cannot be multiplied together r 4 = ( )( ) = 5, B.. [3 marks] x + e k if x If f(x) = x x if x > x is continuous at x = then A. k = B. k = 0 C. k = D. k = E. there is no real number k such that f is continuous at x = lim f(x) = + ek = f() x lim f(x) = lim x x x(x ) = lim x + x + x x + x + e k =, so e k =, so k = 0. B. = lim x =. For continuity, we must therefore have x + Page of 4

16 3. [3 marks] If y = x ln x, then y (e) = A. B. e e C. e D. E. e ln y = ln x ln x = (ln x) y y = ln x x y y ln x = x When x = e, y = e ln e = e = e y = e ln e =. D. e 4. [3 marks] The function f(x) = xe x on the inverval [, ] A. has an absolute minimum at x = and an absolute maximum at x = B. has an absolute minimum at x = and an absolute maximum at x = C. has an absolute minimum at x = and an absolute maximum at x = D. has an absolute maximum at x = and no absolute minimum E. has an absolute minimum at x = and no absolute maximum f (x) = e x xe x = ( x)e x f is continuous on [, ] and has only one critical point, at x =. f must have an absolute maximum and an absolute minimum at x =, x =, or x =. f( ) = e, f() = e, f() = e < e (to see this, start with < e, and divide both sides by e ). This means f() is the maximum and f( ) is the minimum. B. Page 3 of 4

17 5. [3 marks] If the demand equation is p = q3 50q q, for what values of q is the marginal revenue ( dr dq ) decreasing? A. (0, 00) B. (00, 00) C. (00, ) D. All values E. None r = pq = q4 50q3 + 0, 000q dr dq = q3 3 50q + 0, 000q For dr dq to decrease, we need its derivative to be negative. d r dq = q 300q + 0, 000 < 0 (q 00)(q 00) < 0 r (, 00) + (00, 00) - (00, ) + So (00, 00). B. 6. [3 marks] e x x x is lim x 0 x 3 A. undefined B. 0 C. D. 6 E. 4 3 This limit has the indefinite form 0 0 so L Hopital s method applies. e x x x lim x 0 form 0 0 ) = lim x 0 x 3 8e x 6 = lim x 0 e x 4x 3x = 8 6 = 4 3. E. (note this is another 0 0 4e x 4 form) = lim x 0 6x (again of the Page 4 of 4

18 7. [3 marks] If Newton s method is used to estimate a solution to the equation beginning with x =, then x is closest to A B..33 C..39 D..4 E..43 Let f(x) = (x ) ln x f (x) = (x ) x f() = ln = f () = ( ) = 3 x = x f(x ) f (x ) = 3 = + 3 = 4 3. B. (x ) = ln x 8. [3 marks] e ln x x dx = A. B. C. D. e E. e Let u = ln x, du = dx x e ln x x dx = 0 udu = u 0 =. C. Page 5 of 4

19 9. [3 marks] x + x dx = A. (x + ) ln x + C 5 B. ln x + C C. [ln x + ln x + ln x ] + C D. [ln x ln x + ] + C E. [3 ln x ln x + ] + C Using partial fractions: x + x = A x + B x + A(x + ) + B(x ) = x +. Method : When x =, we have A = 3 so A = 3/. When x = we have B = so B = /. Method : (A + B)x + (A B) = x +, so A + B = and A B =. Solve this system of equations for A = 3/, B = /. [ x + 3 Either way, x = x ] dx x + = 3 ln x ln x + + C. E. 0. [3 marks] A town s population is expected to grow exponentially so that it will double every 50 years. population today is 0,000, then in 0 years its population will be A., 039 B., 755 C. 0, 8 D., 487 E. 0, 43 P (t) = 0000e kt 0000 = P (50) = 0000e 50k e 50k = If its e k = 50 P (t) = 0000( t 50 ) P (0) = 0000( 5 ) 487. D. Page 6 of 4

20 . [3 marks] If z = (x + y ) ln(3x + y), then when x = 0 and y = e, A. 0 B. C. e D. e E. 3e z x = x ln(3x + y) + x + y 3x + y 3 when x = 0 and y = e z x = e 3 = 3e. E. e z x =. [3 marks] If y + z = xyz +, then when x = y = z =, A. B. 0 C. D. E. Keeping x constant, y + z z y At x = y = z =, + z z y =. A. y = + z y = xz + xy z y. z y = Page 7 of 4

21 3. [3 marks] If x = 3r + 5s, y = r + 3s, and z = x + y, then when r = s =, A. 64 B. 0 C. 4 D. 68 E. 39 z y = z x x x r + z y y r y z z = 3, =, = x = 6 when r = s =, r r x y z = = 68 when r = s =. D. r z r = = y = 0 when r = s =. 4. [3 marks] The quantities q A and q B of goods A and B, which are sold when the respective unit prices of the goods are p A and p B, are q A (p A, p B ) = 00 5p A + 0p B p B q B (p A, p B ) = 00 6p A 3p B + p A The two goods are complementary A. when p A > 3 and p B < 5 B. when p A > 3 and p B > 5 C. when p A < 3 and p B > 5 D. for no values of p A and p B E. when p A < 3 and p B < 5 A and B are complementary when q A < 0 and q B < 0 p B p A 0 p B < 0 and 6 + p A < 0 p B > 5 and p A < 3. C. Page 8 of 4

22 5. [3 marks] The critical point of is A. (, ), which is a saddle point B. (, ), which is a local max C. (, ), which is a local min f(x, y) = x + xy y x 6y + 3 D. (, ), but the second derivative test cannot determine its nature E. not defined, because the partial derivatives are never both zero f x = x + y = 0 f y = x y 6 = 0 First adding then subtracting these equations, we get 4x 8 = 0 x = 4y + 4 = 0 y = f xx =, f yy =, f xy = f yx = So (, ) is indeed critical. D = f xx f yy (f xy ) = 4 4 = 8 < 0 (, ) is a saddle point. A. Page 9 of 4

23 PART B. WRITTEN-ANSWER QUESTIONS B. [3 marks] At the end of December, 000, Henry invested $00,000 in an annuity which could be refunded at any time and which made 0 equal semiannual payments at the ends of June and December from 00 to 00 inclusive at 6% compounded semi-annually. (a) [4 marks] What was the amount of each payment Henry received from the annuity? Let R = semi-annual payment, i =.03 interest per half-year. Then 00, 000 = Ra , So, R = = $ (.03) 0 (b) [4 marks] At the end of 007, just after the annuity s 4 th payment, Henry refunded it, that is, sold it back to the bank for its current value (value at the end of 007). How much money did he receive? He recieves the principal outstanding, namely, the value of the 6 remaining payments: Ra 6.03 = [ (.03) 6 ] = $36, 4.03 (c) [5 marks] At the beginning of 008, Henry bought as many Acme bonds as he could with the money he received from the sale of the annuity. An Acme bond had a face value of $000, 6 semiannual coupons worth $30 each until maturity at the end of 00, and an annual yield rate of 7%. How many Acme bonds could Henry buy? P = V ( + i) n + rv a n i = 000(.035) a 6.035, the price of a single bond. P = $ per $000 of face value. 36, 4 = So Henry can buy 37 bonds Page 0 of 4

24 B. [ marks] Given: At the present time, 800 people pay $9 each to take a commuter train to work. The number of people willing to ride the train at a price p is given by: (a) [4 marks] At present, is the demand elastic or inelastic? η = = p dq q dp = p q 600( p p ) = 300 q dq q dp p When p = 9 and q = 800, η = = 800 Demand is inelastic q = 600(6 p) (b) [ marks] By examining dr dp revenue (r). r = pq = 600(6p p 3/ ) dr dp = 600(6 3 p ). When p = 9 at the present time, determine whether an increase in price will increase or decrease dr dp = 600(6 9 ) = 900, so an increase in price will increase revenue. (c) [5 marks] Find the price that should be charged in order to maximize revenue. dr dp = 0 when 6 = 3 p/ p / = 4 p = $6. dr dp = 300(4 p/ ), so d r dp = 50p < 0 for all p. So dr dp = 0 gives the maximum. (Or: dr dp > 0 when p < 6, and dr dp < 0 when p > 6). (d) [ mark] At the price found in c), will an increase in price result in an increase or decrease in revenue? Why? An increase in price will decrease revenue, because, at p = 6, revenue is the most it can be. Or, because when p > 6, dr < 0 so revenue decreases as p increases. dp Page of 4

25 B3. [0 marks] (a) [4 marks] Sketch the functions y = x and y = x on the axes below, labelling the points of intersection on the graph Intersection is when x = x. Either x = or, dividing by x (if x ), = x so = x so = x. (b) [6 marks] Shade the region(s) between the graphs of the two functions from x = to x = 5 and evaluate the area of the shaded figure. [Express your final answer as a single number.] [ ] 5 [ ] Area = x (x ) dx + (x ) x dx = [ 3 (x )3/ = 8 6 = 3 ] (x ) + [ (x ) ] 5 3 (x )3/ = [ 3 ] [ + ( ) ( ] 3 ) = Page of 4

26 B4. [0 marks] Find y(x) explicitly in terms of x such that dy dx = xey and y(0) =. [You may assume that x is always between e and e.] dy dx = xey : separate the variables e y dy = xdx e y dy = xdx e y = x + C e y = x + K When x = 0, y = e = K e y = e x y = ln(e x ) (note that we are given x < e so e x > 0). y = ln(e x ) ( or y = ln ). e x Page 3 of 4

27 B5. [0 marks] Use the method of Lagrange multipliers to find all critical points of the problem maximize 4x + y subject to the contraint x + xy = 7 [No marks will be given for any other method.] Make sure to specify the values of the Lagrange multiplier. L = 4x + y λ(x + xy 7) L x = 4 λ(x + y), L y = λx, L λ = (x + xy 7). There are many ways to solve the equations L x = L y = L λ = 0. Here s one: L y = 0 λx =, so λ 0 and x = λ. Then L x = 0 4 λy = 0, so λy = and y = λ. L λ = 0 x + xy = 7, so λ + λ = 7, 3 λ = 7, and we have λ = 9 so λ = ± 3. λ = 3 x = 3 and y = 6. λ = 3 x = 3 and y = 6. Slightly different is: x = λ and y = λ y = x x + xy = 7 3x = 7 x = 9 x = ±3 x = 3 y = 6 and λ = 3 x = 3 y = 6 and λ = 3 Page 4 of 4