f5 c10 linear programing new-1

7/29/2019 f5 c10 Linear Programing New-1

1/8

CHAPTER 10 LINEAR PROGRAMMING FORM 5

PAPER 2

1. Amirah has an allocation of RM200 to buy x workbooks and y reference books. The total

number of books is not less than 20. The number of workbooks is at most twice the number ofthe references. The price of a workbook is RM10 and that of a reference is RM5.

a ) Write down three inequalities, other than x 0 and y 0, which satisfy all the aboveconstraints.

( 3 marks)

b ) Hence, using a scale of 2cm to 5 books on the x-axis and 2cm to 5 books on the y-axis,

construct and shade the region R that satisfies all the above constraints. ( 4 marks )

c ) If Amirah buys 15 reference books, find the maximum amount of money that is left.

( 3 marks )

2. A university wants to organise a course for x medical undergraduates and y dentistry

undergraduates. The method in which the number of medical undergraduates and dentistryundergraduates are chosen are as follows.

I : The total number of participants is at least 30.II : The number of medical undergraduates is not more than three times the number of dentistry

undergraduates.

III : The maximum allocation for the course is RM6 000 with RM100 for a medicalundergraduates and RM80 for a dentistry undergraduates.

a) Write down three inequalities, other than x 0 and y 0, which satisfy all the aboveconstraints.

( 3 marks )

b) Hence, by using a scale of 2cm to 10 participants on both axes, construct and shade the regionR that satisfies all the above constraints. ( 3 marks )

c) Using your graph from (b), find( i) The maximum and minimum number of dentistry undergraduates , if the number of

medical undergraduates that participate in the course is 20.

(ii) The minimum expenditure to run the course in this case. ( 4 marks )

3. A tuition centre offers two different subjects, science, S, and mathematics, M, for Form 4

students. The number of students for S is x and for M is y. The intake of the students is basedon the following constraints.

I : The total number of students is not more than 90.

II : The number of students for subject S is at most twice the number of students for subjectM.

III : The number of students for subject M must exceed the number of students for subject S

by at most 10

143


2/8


a) Write down three inequalities, other than x 0 and y 0, which satisfy all the above

conditions.

( 3 marks)

b) Hence, by using a scale of 2 cm to represent 10 students on both axes, construct and shade the

region R that satisfies all the above conditions. ( 3 marks )

c) Using the graph from ( b ), find ( 4 marks )

( i ) the range of the number of students for subject M if the number of students for subjects S

is 20(ii) the maximum total fees per month that can be collected if the fees per month for subject S

and M are RM12 and RM10 respectively.

4 A bakery shop produces two types of bread, L and M. The production of the bread involves two

processes, mixing the ingredients and baking the breads. Table 1 shows the time taken to make

bread L and M respectively.

Type of bread Time taken ( minutes )

Mixing the ingredients Baking the breads

L 30 40

M 30 30

Table 1

The shop produces x breads of type L and y breads of type M per day. The production of breads

per day are based on the following constraints:

I : The maximum total time used for mixing ingredients for both breads is not more than 540

minutes.II : The total time for baking both breads is at least 480 minutes.

III : The ratio of the number of breads for type L to the number of breads for type M is not less

than 1 : 2

a ) Write down three inequalities, other than x 0 and y 0, which satisfy all the above

constraints. ( 3 marks )

b) Using a scale of 2 cm to represent 2 breads on both axes, construct and shade the region Rthat satisfies all the above constraints. ( 3 marks )

c) By using your graph from 4(b), find

( i ) the maximum number of bread L if 10 breads of type M breads are produced per day.

(ii ) the minimum total profit per day if the profit from one bread of type L is RM2.00 and

144


3/8


from one bread of type M is RM1.00 . ( 4 marks )

5. A factory produced x toys of model A and y toys of model B. The profit from the sales of a

number of model A is RM 15 per unit and a number of model B is RM 12 per unit.The production of the models per day is based on the following conditions:-

I : The total number of models produced is not more than 500.II : The number of model A produced is at most three times the number of model B.III : The minimum total profit for model A and model B is RM4200.

a) Write down three inequalities, other than x 0 and y 0, which satisfy all the aboveconditions.

( 3 marks)

b) Hence, by using a scale of 2 cm to represent 50 models on both axes, construct and shade the

region R that satisfies all the above conditions. ( 3 marks )

c) Based on ypur graph, find ( 4 marks )( i ) the minimum number of model B if the number of model A produced on a particular

day is 100.

(ii) the maximum total profit per day

145


4/8


1. ( a ) I : x + y 20

II : x 2yIII : 10 x + 5y 200

2x + y 40

( b )

( c ) Draw the line y = 15From the graph, the minimum value occurs at ( 5, 15 )

Hence, minimum expenditure = 10 (5) + 5 (15)

= RM125

Therefore, the maximum amount of money left = RM200 RM125=RM 75

1

11

4

11

1

146

4

3

3

2

2

1

1

5

1 2 3 4 5 6

R y = 15

x + y = 20

2x + y = 40

x = 2y


5/8


2.

( a ) I : x + y 30

II : x 3yIII : 100 x + 80y 6000

5x + 4y 300

( b )

( c ) ( i ) Draw the line x = 20,

From the graph, the minimum number of dentistry undergraduates is 10 and themaximum number of dentistry undergraduates is 50.

(ii) For the minumum expenditure, there are 20 medical undergraduates and 10 dentistry

undergraduates.Thus , minimum expenditure = 100 (20) + 80(10)

=RM2800

1

11

3

1

1

1

1

147

9

8

7

6

5

4

3

2

1

-1

2 20 40 60 8 100 120 140

x = 3y

R

5x + 4y = 300

x = 20

x + y =30


6/8


148

3

. ( a ) I : x + y 90

II : x 2y

III : y - x 10

( b )9

8

7

6

5

4

3

2

1

-1

2 40 6 80 100 120 140

12x + 10y = 600

x + y = 90

x = 2y

R

y - x = 10

( c ) ( i ) From the graph, when x = 20, the range of y is 10 y 30

( ii ) The maximum value occurs at ( 60, 30)

Thus, maximum total fee = 12 (60) +10 (30)

= RM1 020

1

1

1

3

1

1

11


7/8


149

4( a ) I : 3x + 3y 54

II : 4x + 3y 48

III : 2x y

( b )22

20

18

16

1

12

10

8

6

4

2

-

5 10 15 20 25 30 35

R

2x + y = 2

4x + 3y = 4 8

3x + 3y =54

y = 2x

( c ) (i ) 8

(ii ) (5 x RM2.00) + ( 10 x RM1.00)= RM 20.00

1

1

1

3

1

2

1


8/8


5 (a ) I : x + y 500

II : x 3y

III : 15x + 12y 42005x + 4y 1400

( b )

( c ) ( i ) When x = 100, the minimum number of model B is 225.

(ii ) Maximum point (375, 125)

The maximum total profit per day= 15(375) +12(125)

= RM7125

1

1

1

3

1

1

1

150

f5 c10 linear programing new-1

Documents