MATA33 Exams Solutions 2008-2017 PDF

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MATA33 Exams Solutions 2008-2017
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MAT A33 - CALCULUS FOR MANAGEMENT II UTSC Year 1 Course

FINAL EXAMS - SOLUTIONS years 2008 - 2017

FULL SOLUTIONS for 16 Final Exams with How-To explained in detail

Contents

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16

Introduction

……………………………………………………………..

3

Final Exam 2017, Apr 21 Final Exam 2016, Aug 15 Final Exam 2016, Apr 8 Final Exam 2015, Apr 22 Final Exam 2014, Aug 13 Final Exam 2014, Apr 16 Final Exam 2013, Apr 19 Final Exam 2012, Aug 22 Final Exam 2012, Apr 24 Final Exam 2011, Aug 13 Final Exam 2011, Apr 20 Final Exam 2010, Aug 13 Final Exam 2010, Apr 15 Final Exam 2009, Aug 19 Final Exam 2009, Apr 24 Final Exam 2008, Apr 22

…………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. …………………………………………………………….. ……………………………………………………………..

4 16 30 43 56 69 83 98 111 123 137 149 159 171 184 200

Introduction This book contains fully explained solutions to all questions from the Final Exams for MAT A33 - Calculus for Management 1I, University of Toronto Scarborough Campus first year course, from the year 2008 to 2017. All answers provide fully explained How-To details, often with additional explanations and graphs that are not required by the examiner but are provided to improve an understanding of the answer. Answers are provided to all questions (252 in total) for each of the MAT A32 final exams (16 in total) over the 2008 through 2017 time period. All questions provided in the book can also be found on the MAT A32 website

https://www.math.utsc.utoronto.ca/a33/assignments.html To help the reader to connect with the topic/problem at hand and when it deemed appropriate, a reference is given to the Textbook used for the MAT A32 course: “Introductory Mathematical Analysis for Business, Economics and The Life and Social Science”, 13th Edition, ISBN-13:978-0-321-64372-8, by E.F. Haeussler, Jr., R.S. Paul and R.J. Wood. Readers’ comments, positive and/or negative, about any question/answer, will be much appreciated.

In Markham, Ontario, March 2019. Greg Crnilovic, Math Tutor [email protected]

2008_Apr 22

Part A: Multiple Choice Questions: For each of the following print the letter of the answer you think is most correct in the boxes on the first page. Each right answer earns 4.5 points and no answer/wrong answers earn 0 points. Justification is neither required nor rewarded, but a small workspace is provided for your calculations. 1.

x+y then fx(2, 1) equals If f(x, y) = x-y

(a)

(b) 2

-2

1 (x - y) - (x (x-y)2 fx(2, 1) = - 122 = -2

fx =

(c)

1

d(0)

2y + y) = - (x-y) 2

⟹ Correct answer is (a).

2.

At the point (1, 1) the function f(x, y) = x 4 + y4 has (a) a relative minimum (b) a relative maximum

(c)

a saddle point

(d) none of the above We look for the critical point(s) first: fx = 4 x 3 = 0 fy = 4 y3 = 0

⟶ x=0 ⟶ y=0

Function f has one critical point only, at (0, 0). Since the critical point is the only point on the curve for relative extrema (maximum or minimum) and saddle point, given point (1, 1) cannot be either of the three.

⟹ Correct answer is (d).

3.

Assume A and B are 2 x 2 matrices such that A 3 - AB2 = 0. Then (a) none of (b), (c) or (d) need be true (b) A= B or A = -B or A = 0 (c) A 2 = B2

Let A = 

0 0  1 0

A 2 = AA = 

and

B=

0 0  1 1

0 0 0 0 0 0 0 0 0 0 0 0  ⟶ A 3 = AA·A =   · · = = 0 0 0 0 1 0 0 0 1 0 1 0

0 0 0 0 0 0 · =  1 1 1 1 1 1 0 0 0 0 0 0 AB2 =  · =  1 0 1 1 0 0 B2 = BB = 

(d) AB = BA

2008_Apr 22

A 3 - AB2 = 

0 0 0 0 0 0 - =  0 0 0 0 0 0

as required.

Now we get: AB = 

0 0 0 0 0 0 · =  1 0 1 1 0 0

BA = 

0 0 0 0 0 0  · = 1 0 1 1 1 0

Therefore: A ≠ B and A ≠ -B and A ≠ 0 A 2 ≠ B2 eliminates choice (c) AB ≠ BA eliminates choice (d)

eliminates choice (b)

⟹ Correct answer is (a).

4.

2 6 ∫1 ∫3 x2 dydx is equal to 15 4

(a)

(b)

7

(c)

45 4

(d) 63

I = ∫1 ∫3 x 2 dy dx = ∫1 x 2(6 - 3) dx = 3· 13 23 - 13) = 7 2

6

2

⟹ Correct answer is (b).

5.

The domain of the function f(x, y) = (1 - x 2 - y2)-1/2 is exactly (a) the set of points that are strictly outside the unit circle (b) the set of points that lie on or outside the unit circle (c) the set of points that lie on or inside the unit circle (d) the set of points that are strictly inside the unit circle y

1.0

0.5

0.0

x

-0.5

-1.0 -1.0

-0.5

0.0

0.5

1.0

f (x, y) = (1 - x 2 - y2)-1/2 =

1 1-x2 +y2 

The domain is where 1 - (x 2 + y2) ≥ 0

⟶ x 2 + y2 ≤ 1

which means that given set of points lies on and inside the unit circle x 2 + y2 = 1 as shown in the

2008_Apr 22

diagram above.

⟹ Correct answer is (c).

6.

A general system of m linear equations in n unknowns where n > m > 1 (a) has at least one solution (c) has a unique solution

(b) has infinitely many solutions (d) may not have any solution

⟹ Correct answer is (d).

Note: Generally, in the case of systems with more unknowns then equations, system can be either consistent with infinitely many solutions (rightmost column of the augmented matrix is not a pivot column) or inconsistent (rightmost column of the augmented matrix is a pivot column), i.e. with no solution at all.

7.

If z = x 2 exy and x and y are independent variables, then (a) exy(2 + xy) (c) 2 xyexy

∂z (x, ∂x xy

y) equals

(b) xe (2 + xy) (d) xexy(2 + x 2)

zx = 2 xexy + x 2 yexy = xexy(2 + xy) ⟹ Correct answer is (b).

8.

The equation of the level curve of h(x, y) = (a)

y = -1

2

(b) y = -x - 1

y x2 +1

- 2 that passes through (2, 5) is

(c)

y = x2 + 1

(d) none of the above

We first find the value of the function at the given point: h(2, 5) =

5 22 +1

- 2 = -1

For the level curve at (2, 5) we have h(x, y) = x2y+1 - 2 = -1 ⟶ y = x2 + 1 ⟹ Correct answer is (c).

9. The joint demand functions for two products X and Y are given by f(x, y) = 20 - x - e -y and g(x, y) = 50 + x + 3 y where x and y are the unit prices of X and Y, respectively. We may conclude that the products are (a) complementary (b) competitive (c) both (a) and (b) (d) neither (a) nor (b) fy = gx =

∂ f(x, y) = e-y > 0 ∂y ∂ g(x, y) = 1 > 0 ∂x 2 x

for y > 0 for x > 0 (x, y > 0 - unit prices)

2008_Apr 22

As both partial derivatives above are positive, two products X and Y are competitive products.

⟹ Correct answer is (b).

10. Let R be the feasible region consisting of all points in and on the sides of the triangle with corners (0, 0), (2, 0) and (1, 1). If z = ax + by where a and b are non-zero constants then we may conclude that (a) z has a maximum at a corner point (b) there exist values of a and b for which z has a maximum at every point on some side of R (c) the minimum value of z occurs at (0, 0) (d) only (a) and (b) are true (e) each of (a), (b) and (c) is true y

Q (1,1)

1

R

S (0,2)

P (,0,0)

1

2

x

We have to analyze the value of Z = ax + by at the corner points of the triangle PQS: P(0, 0) Q(1, 1) S(2, 0)

⟶ z = a ·0 + b·0 = 0 ⟶ z = a ·1 + b·1 = a + b ⟶ z = a ·2 + b·0 = 2 a

Three scenarios have to be considered: (i) both a and b are negative non-zero constants: z is negative everywhere except at the origin

⟹ z(0, 0) = 0 = zmax

⟶ assumption (a) is true , and assumption (c) is false. (ii) both a and b are positive non-zero constants: z(0, 0) = 0 z(1, 1) = a + b > 0 if a < b then z(1, 1) = zmax z(2, 0) = 2 a > 0 if a > b then z(2, 0) = zmax -

z(1, 1) = z(0, 2) = 2 c > 0 if a = b = c then z(1, 1) = z(0, 2) = zmax

⟶ assumption (a) is true , and assumption (c) is true.

2008_Apr 22

(iii) We explore if the assumption (c) might be true along any of the three sides. Let’s consider the case when both a and b are positive and equal, a = b = c > 0, and check the value of z along 3 sides: z = ax + by = cx + cx = 2 cx ⟶ Z is not a constant (increasing as c, x > 0) Z = cx + 0 = cx ⟶ Z is not a constant (increasing as c, x > 0) - QS (y = 2 - x) Z = cx + c(2 - x) = 2 c ⟶ Z is positive and constant. Because 2 cx ≤ 2 c for x on [0,1) on PQ and also cx ≤ 2 c for x on [0,2] on PS we conclude - PQ (y = x) - PS (y = 0)

that z has a maximum along the QS side

⟶ assumption (b) is true.

As per (i) - (iii) above, only the assumptions (a0 and (b) are always true. Correct answer is (d).

2008_Apr 22

Part B: Full-Solution Questions. Write clear and neat solutions in the answer spaces provided. Show all of your work. Full points are awarded for solutions only if they are correct, complete, and sufficiently display relevant concepts from MATA33. 1. Use the method of Lagrange multipliers to find the maximum value of f(x, y) = xy + 2 x subject to the constraint 2 x + y = 30. (You may assume that the critical point obtained does correspond to a maximum).

[7 points]

g(x, y) = 2 x + y - 30 = 0 F(x, y, λ) = f(x, y) - λg(x, y) = xy + 2 x - λ(2 x + y - 30) Fx = y + 2 - 2 λ = 0

Fy = x - λ = 0 Fλ = -(2 x + y - 30) = 0

(1) & (2) ⟶ y - 2 x + 2 = 0 (4) & (3) ⟶ -4 x + 32 = 0 ⟹

x=8

...

(1)

... ...

(2) (3)

...

(4)

y = 14

f(8, 14) = fmax = 8·14 + 2·8 = 128

2.

2

A joint cost function is given by c(x, y) = ln(xy2 + 1) + 6yx where x and y represent the numbers

of units of two products. (a) Find the marginal cost functions at (4, 3) rounded to 2 decimal places. y2

12 x y 32 4·32 +1 2·4·3 4·32 +1

cy = xy22xy+1 -

cx = xy2 +1 + cx(3, 5) = cy (4, 3) =

+ -

12·4 3 6·42 32

[7 points]

6 x2 y2

= 16.24 = -10.02

(b) State a mathematical relationship that involves the cost function at (4, 3) and (5, 3), and a marginal cost function at (4, 3).

[3 points]

We check the real cost increase when products’ output changes from (4, 3) to (5, 3). We observe that change occurs in x-output only (from 4 to 5 x-products produced): c(x, y) = ln(xy2 + 1) + 6 yx

2

2

c(4, 3) = ln(4·32 + 1) + 6·43 = $35 .61 2 c(5, 3) = ln(5·32 + 1) + 6·53 = $53 .83

2008_Apr 22

The real cost increase is a deference, Δc 4 to 5 = 53.83 - 35.61 = $18 .22 Since the change, from (4, 3) to (5, 3), occurred in x-output only, we compare Δc to marginal cost ∂c = cx found in (a) above. ∂x

We confirm that cx = $16 .24 is a good approximation of the real cost change (increase!) of

Δc = $18 .22, when x-output is changed from 4 to 5.

Let f(x, y) = 3 xey - x 3 - e3 y . Find the critical point(s) of f. For each one, use the second derivative test to determine whether it corresponds to a relative maximum, minimum, or a saddle point. State the maximum/minimum value(s) of f should they occur. [10 points] 3.

f(x, y) = 3 xey - x 3 - e3 y fx = 3 ey - 3 x 2 = 0 fy = 3 xey - 3 e3 y = 0

⟶ x 2 = ey ⟶ x = e 2 y = (ey )2

...

... (1) (2)

x = (x 2)2 = x 4 ⟶ x 4 - x = 0 ⟶ x(x 3 - 1) = 0 (i) x = 0 ⟶ ey = 0 ⟹ y = ∞ which means that x = 0 can be ignored. (ii) x = 1 ⟶ ey = 1 ⟹ y = 0 which means that (1, 0) is the critical point of f.

(2) & (1)

fxx = -6 x y

fyy = 3 xe - 9 e fxy = 3 e

fxx (1, 0) = -6 3y

fyy(1, 0) = 3·1 - 9·1 = -6

y

D(x, y) = fxx fyy - fxy

fxy(1, 0) = 3·1 = 3 2

D(1, 0) = (-6)·(-6) - 32 = 27 > 0

With D > 0 and fxx < 0 we get that f has a relative maximum at critical point (1, 0) and its value is f(1, 0) = fmax = 3·1 - 1 - 1 = -1 1 0 0 0 0 -1 0 0 4. Let A = 0 0 2 3 0 0 3 -6 invertible. [10 points] 1 0 0 0 0 -1 0 0 B(x) = A - xI = -x 0 0 2 3 0 0 3 -6 1-x 0 0 0 0 -1 - x 0 0 B(x) = 0 0 2-x 3 0 0 3 -6 - x

Find all real values of x for which the matrix B(x) = A - xI is not

1 0 0 0

0 1 0 0

0 0 1 0

0 1 0 0 0 x 0 0 0 x 0 -1 0 0 = 0 0 0 2 3 0 0 1 0 0 0 0 3 -6

0 0 x 0

0 0 0 x

2008_Apr 22

2-x 3  = (x - 1) (x + 1)[(2 - x) (-6 - x) - 9] 3 -6 - x det B(x) = (x - 1) (x + 1)[(x - 2) (x + 6) - 9] = (x - 1) (x + 1) (x 2 + 4 x - 21) det B(x) = (x - 1) (x + 1) (x + 7) (x - 3) (x + 7) (x - 3) det B(x) = (1 - x) (-1 - x)·det 

Matrix is invertible if its determinant is equal to zero: det B(x) = (x - 1) (x + 1) (x + 7) (x - 3) (x + 7) (x - 3) = 0 which is true for x = {1, -1, -7, 3} .

5.

let z =

x y

where x = set and y = 1 + se-t . Use the chain rule to find

∂z ∂s

when s = 2 and t = 0.

[7 points] The chain rule formula for partial derivatives (Textbook, p. 765) gives ∂x ∂z ∂y ∂z = ∂z + ∂y ∂s let z = yx where x = set and y = 1 + se-t s=2 ∂s ∂x ∂s

t=0

x(2,0) = 2

y(2,0) = 1+2 = 3 zx =

1 y

zy = - yx2

xs = et

zs = zx xs + zy ys =

1 y

et - yx2 e-t

ys = e-t

and for {x, y, s, t} = {2, 3, 2, 0}

⟹

∂z ∂s

6.

Assume the equation z3 + 2 x 2 z2 = xy defines z implicitly as a function of independent

= zs =

1 ·1 - 29 ·1 = 19 3

variables x and y. Find zx when x = z = 1. z3 + 2 x 2 z2 = xy

[7 points]

∂ ∂x

3 z2 ·zx + 2 (2 xz2 + x 2 ·2 z·zx) = y zx(3 z2 + 4 x 2 z) + 4 xz2 = y xz2 zx = 3 zy-4 2 +4 x2 z x =z=1

y=

1+2·1 =3 1 1 3-4·1 1) = 3·1+4·1 = -7

z3 +2 x2 z2 x

⟹ zx(x, y, z) = zx(1, 3,

=

7. A company makes two products, A and B, whose selling prices per unit are $x and $y, respectively. Their fixed production costs are $m and $n per unit, respectively. The demand for A is f(x, y) = 5 (y - x ) and the demand for B is g(x, y) = 500 + 5 (x - 2 y). Find the selling prices that maximize the total profit and verify that these prices actually do maximize profit. (A useful concept is that Total Profit = Total Revenue - Total Cost). [12 points] Let Profit, Revenue and Cost be P, R and C respectively. RA = x ·f(x, y) CA = m ·f (x, y)

2008_Apr 22

⟶ PA = RA - CA = (x - m)·f(x, y) = 5 (x - m) (y - x] = 5 (xy - x 2 - my + mx) RB = y· g(x, y) CB = n · g(x, y) ⟶ PB = (y - n)· g(x, y) = 5 (y - n)[100 + (x - 2 y] = 5 (100 y + xy - 2 y2 - 100 n - nx + 2 ny) P = PA + PB = 5[2 xy - x 2 - 2 y2 + (m - n) x + (2 n - m + 100) y - 100 n]

Px = 5 (2 y - 2 x + m - n) = 0 ⟶ -2 x + 2 y = -m + n Py = 5 (2 x - 4 y + 2 n - m + 100 = 0 ⟶ 2 x - 4 y = m - 2 n - 100 (1) + (2) →

-2 y = -n - 100

⟶

(1) →

-2 x + 2· 12 (n + 100) = -m + n ⟶

y=

1 2

... ...

(1) (2)

(n + 100) x = 12 (m + 100)

8. Let a and b be constants such that 0 < b < a. Use techniques from linear programming to find the value of a and b so that the objective function z = ax + by has a maximum value of 46 and a minimum value of 14 when (x, y) is subject to the constraints 3 ≤ x ≤ 9, x + 3 y ≥ 6, x - 3 y ≥ -6 and x , y ≥ 0. [12 points] C(9, 5)

5

x - 3 y ≥ -6

4

3

B (3, 3)

Feasable Region is inside and along the sides of the pentagon ABCDE.

x=9 2

x=3

1

A(3, 1)

D (9, 0)

E(6, 0) 0

3

4

5

6

7

8

9

Bearing in mind that a > b (given condition), we look for the maximum at the points where in the expression ax + by, ‘x-coefficient’ takes maximum value. This happens at the points C(9, 5) and D(9, 0) . Point C(9, 5), where z(9, 5) = 9 a + 5 b is the clear choice here, because z(9, 5) = 9 a + 5 b > z(9, 0) = 9 a.

2008_Apr 22

Similar reasoning leads to the conclusion that the minimum should be looked for at the point A(3, 1) with z(3, 1) = 3 a + b. So, for zmax = 46 and zmin = 14, we end up with two equations 9 a + 5 b = 46 ... (1) 3 a + b = 14 ... (2) a=

9.

46 14 9 det 3

det

5  1 5  1

=

46-70 9-15

= 4 and

b=

9 3 9 det 3

det

(a) Evaluate I = ∫1 ∫0ln(x) dydx e x

I = ∫1 ∫0 [ln(x) dy] dx = ∫1 [ln(x )· y e x

e

1 x

u = ln(x)

=

126-138 9-15

=2

[9 points]

e x 0 ] dx = ∫1 x ln(x) dx

∫ udv = uv - ∫ vdu

du = dx

xdx = dv v = 12 x 2 e I = 12 x 2 ln(x) - ∫1 21 x 2 · 1x dx

I = 12 x 2 ln(x) - 21 · 12 x 2 e1 = 14 [2 x 2 ln(x) - x 2] I = 14 [2 e2 ·1 - e2 - (0 - 12)] = 41 (e2 + 1)

10.

46  14 5  1

e 1

A rectangular box with no top has a volume of 60 m3. The material for the front and bottom costs

$5/m2 and the material for the back, and left and right sides costs $1 /m2. Use the method of Lagrange multipliers to find the dimensions of the box whose material cost is smallest (You may assume that the critical point produced actually does result in a least material cost). [14 points]

z

x

y

V = xyz = 60 C = 5 (xz + xy) + 1 (xz + 2 yz) = 5 xy + 6 xz + 2 yz

g(x, y, z) = xyz - 60 = 0

F(x, y, z, λ) = C(x, y, z) - λg(x, y, z) F(x, y, z, λ) = 5 xy + 6 xz + 2 yz - λ(xyz - 60) Fx = 5 y + 6 z - λyz = 0 Fy = 5 x + 2 z - λxz = 0 Fz = 6 x + 2 y - λxy = 0 Fλ = -(xyz - 60) = 0

... ... ...

(1) (2) (3)

...

(4)

2008_Apr 22

⟶ 5 xy + 6 xz - λxyz = 0 ⟶ 5 xy + 2 yz - λxyz = 0 ⟶ 6 xz + 2 yz - λxyz = 0

(1)*x (2)*y (3)*z (1-a) & (2-a) (2-a) & (3-a) (1-a) & (3-a)

... ... ...

(1-a) (2-a) (3-a)

⟶ 6 xz - 2 yz = 0 ⟶ 3 x = y (z ≠ 0) ⟶ 6 xz - 5 xy = 0 ⟶ 6 z = 5 y (x ≠ 0) ⟶ 5 xy - 2 yz = 0 ⟶ 5 x = 2 z (y ≠ 0)

(4)

⟶ x ·3 x · 2 x = 60

⟶

x=2

(1-a)

⟶ 3·2 = y

⟶

y=6...