MTH-5101 Pretest A Solutions
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Mth 5101 pretest a solutions

MTH-5101 Pretest A Solutions

  • A sport boutique sells two types of golf balls: standard quality at $2 each and high quality at $3.50 each. The shopkeeper notices that on a monthly basis he sells at least 2 times more standard quality balls than high quality ones. The shopkeeper ensures that he always has at least 50 standard quality and 25 high quality balls in stack.. However, he never has more than 150 in total. How many balls of each kind must he sell to maximize his revenues?

  • a) Transcribe the elements needed to establish the constraints.

  • b) Transcribe the elements needed to establish the function to be optimized.

  • A sport boutique sells two types of golf balls: standard quality at $2 each and high quality at $3.50 each. The shopkeeper notices that on a monthly basis he sells at least 2 times more standard quality balls than high quality ones. The shopkeeper ensures that he always has at least 50 standard quality and 25 high quality balls in stack.. However, he never has more than 150 in total. How many balls of each kind must he sell to maximize his revenues?

  • a) Transcribe the elements needed to establish the constraints.

  • b) Transcribe the elements needed to establish the function to be optimized.

  • A sport boutique sells two types of golf balls: standard quality at $2 each and high quality at $3.50 each. The shopkeeper notices that on a monthly basis he sells at least 2 times more standard quality balls than high quality ones. The shopkeeper ensures that he always has at least 50 standard quality and 25 high quality balls in stack.. However, he never has more than 150 in total. How many balls of each kind must he sell to maximize his revenues?

  • a) Transcribe the elements needed to establish the constraints.

  • b) Transcribe the elements needed to establish the function to be optimized.

  • Yves is the owner of an arena. He wants to improve the lighting on the skating rink. To do the work, he hires an electrician and an apprentice. He evaluates the time needed to do the work at a maximum of 50 hours. The apprentice must work at least 12 hours and the electrician must work at least as many hours as the apprentice.

  • The material required to complete the job costs $10500. The electrician is paid $30 an hour and the apprentice is paid $18 an hour.

  • a) Identify the variables.

  • b) Translate the constraints into a system of inequalities.

  • c) Express the function to be optimized.

Let x = the number of hours worked by electrician

Let y = the number of hours worked by apprentice

y ≥ 12

x ≥ y

x + y ≤ 50

x ≥ 0

y ≥ 0

30x + 18y + 10500 = Z


Mth 5101 pretest a solutions

30

C5: 5x – 8y = 20

20

10

18

0

x

36

x

4

y

20

0

25

4

y

10

20

30

C6: 7x + 6y = 150

DIMENSION 3

3. Draw the polygon of constraints associated with the following system of inequalities.

C1: x ≥ 0

C4: y ≥ 5

C2: y ≥ 0

C5: 5x - 8y ≤ 20

C3: x ≥ 2

C6: 7x + 6y ≤ 150

10


Mth 5101 pretest a solutions

4. Verify algebraically whether the point (25,25) belongs to the polygon of constraints. Show all steps to your solution.

x ≥ 0

25 ≥ 0 True

y ≥ 0

25 ≥ 0 True

x ≥ 20

25 ≥ 20 True

y ≥ 10

25 ≥ 10 True

x + y ≤ 50

25 + 25 ≤ 50

50 ≤ 50 True

6x + 12y ≥ 300

6(25) + 12(25) ≥ 300

150 + 300 ≥ 300

450 ≥ 300 True

The point (25,25) belongs to the polygon of constraints because it makes all of the constraints true.


Mth 5101 pretest a solutions

400

300

x + y = 150

x = 2y

2y + y = 150

x + y = 150

y = 0

x + 0 = 150

x = 150

200

3y = 150

y = 50

C

x = 2y

x = 2(50)

x = 100

100

B

D

A

x = 2y

x + y = 450

2y + y = 450

3y = 450

y = 150

100

200

300

400

x + y = 450

y = 0

A

x + y = 150

y = 0

(150,0)

x + 0 = 450

x = 450

B

x + y = 150

x = 2y

(100,50)

x = 2y

x = 2(150)

x = 300

C

D

x + y = 450

x = 2y

x + y = 450

y = 0

(450,0)

(300,150)

5. Determine algebraically the coordinates of the vertices of the polygon of constraints.

C1: x ≥ 0

C4: x + y ≥ 150

C3

C2: y ≥ 0

C5: x ≤ 2y

C3: x + y ≤ 450

C5

C4


Mth 5101 pretest a solutions

B

x + y = 500

A

400

x = 150

y = 0

0

500

x

0

400

x

y

500

0

400

y

0

x = y

x + y = 500

y + y = 500

2y = 500

y = 250

300

x = 150

x = y

150 = y

y = 150

y = 0

x + y = 500

x + 0 = 500

x = 500

x = y

200

C

x = y

x = 250

100

D

100

200

300

400

6. A firm puts both a shampoo and a conditioner on the market. Both products are sold in 500-ml bottles. A study determined the following information: each month the number bottles of shampoo sold will be at least as many as that of the conditioner. The firm will sell a total of at most 500 bottles in a month and at least 150 of those will be shampoo. After a market survey, the shampoo was priced at $3.80 per bottle and the conditioner at $3.10 per bottle. How many bottles of each kind must the firm sell to maximize its revenue? What would be that revenue? Show all of your work.

Let x = the number of bottles of shampoo sold

Let y = the number of bottles of conditioner sold

x ≥ 0

x ≥ y

y ≥ 0

x ≥ 150

3.8x + 3.1y = Z

x + y ≤ 500


Mth 5101 pretest a solutions

B

A

400

300

A

x = 150

y = 0

(150,0)

200

B

x = 150

x = y

(150,150)

C

100

D

C

x + y = 500

y = 0

x + y = 500

x = y

(250,250)

(500,0)

D

100

200

300

400

Let x = the number of bottles of shampoo sold

Let y = the number of bottles of conditioner sold

3.8x + 3.1y = Z

A(150,0)

3.8x + 3.1y = 3.8(150) + 3.1(0) = 570 + 0

= $570

B(150,150)

3.8x + 3.1y = 3.8(150) + 3.1(150) = 570 + 465

= $1035

C(250,250)

3.8x + 3.1y = 3.8(250) + 3.1(250) = 950 + 775

= $1725

D(500,0)

3.8x + 3.1y = 3.8(500) + 3.1(0) = 1900 + 0

= $1900

To maximize revenue the firm must sell 500 bottles of shampoo and 0 bottles of conditioner. Its revenue would then be $1900.


Mth 5101 pretest a solutions

x + y = 90

y = x

y = x

x + y = 90

x + x = 90

2x = 90

x = 45

y = 45

A

80

x = 10

y = 20

y = 20

y = x

20 = x

x

x

0

0

80

90

y

90

0

80

0

y

60

B

x = 10

x + y = 90

10+ y = 90

y = 80

40

C

D

20

20

40

60

80

7. John makes 2 types of badges: sport badges and ecological badges. The sport badges cost $9 each to make whereas the ecological badges cost $13. John can’t produce more than 90 badges. He must produce at least 10 sport badges and 20 ecological badges. He must produce at least as many ecological badges as sport badges. His advertising campaign costs $100. How many badges of each type must he produce to minimize his costs. What would be that cost? Show all of your work.

Let x = the number of sport badges

Let y = the number of ecological badges

9x + 13y + 100 = Z

x ≥ 0

x ≥ 10

y ≥ 0

y ≥ 20

x + y ≤ 90

y ≥ x


Mth 5101 pretest a solutions

A

80

60

B

A

x = 10

y = 20

(10,20)

40

B

x + y = 90

x =10

(10,80)

D

C

C

20

D

x + y = 90

y = x

y = x

y = 20

(20,20)

(45,45)

20

40

60

80

Let x = the number of sport badges

Let y = the number of ecological badges

9x + 13y + 100 = Z

A(10,20)

9x + 13y + 100 = 9(10) + 13(20) + 100 = 90 + 260 + 100

= $450

B(10,80)

9x + 13y + 100 = 9(10) + 12(80) + 100 =90 + 1080 + 100

=$1270

C(45,45)

9x + 13y + 100 = 9(45) + 13(45) + 100 =405 + 585 + 100

=$1090

D(20,20)

9x + 13y + 100 = 9(20) + 13(20) + 100 =180 + 260 + 100

=$560

10 sport badges and 20 ecological badges are necessary to minimize costs.


Mth 5101 pretest a solutions

(2,7)

(8,4)

(2,4)

  • At the district playground, the instructors planned an outdoor activity at the Bromont slides. Buses had to be rented to transport between 120 and 240 children. Two kinds of buses were available: 15-passenger and 30-passenger. Each small bus cost $40 and each big bus cost$50.

  • The polygon of constraints is presented in the graph on the right where:

8

(2,7)

x = the number of 15-passenger buses

y = the number of 30-passenger buses

6

The instructors later learn that they must rent at least 4 30-passenger buses. How will this effect the decision on minimizing the cost of the outdoor activity.

4

Function of Optimization: 40x + 50y = Z

(2,3)

(2,3)

40x + 50y = 40(2) + 50(3) = 80 + 150

= $230

2

(2,7)

40x + 50y = 40(2) + 50(7) = 80 + 350

= $430

(6,1)

(14,1)

(6,1)

40x + 50y = 40(6) + 50(1) = 240 + 50

= $290

(14,1)

40x + 50y = 40(14) + 50(1) = 560 + 50

= $610

4

8

12

16

According to the polygon of constraints presented, the minimal cost to pay would be $230. With the additional consideration, the minimal cost would have to be higher because with 2 15-passenger buses and 3 30-passenger buses, they will NOT have 4 30-passenger buses.

(2,7)

40x + 50y = 40(2) + 50(7) = 80 + 350

= $430

y ≥ 4

(2,4)

40x + 50y = 40(2) + 50(4) = 80 + 200

= $280

(8,4)

40x + 50y = 40(8) + 50(4) = 320 + 200

= $520

To be consistent with the additional consideration, they would have to use 2 15-passenger buses and 4 30-passenger buses at a cost of $280. This has the effect of increasing the minimum cost from $230 to $280.


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