SARM@N'S ( AIS 25th Batch )
Linear programing problem
Previous – 01 : 22, 20 (Class)
A company makes two kinds of products: Model I and Model II. Each product of Model I sells for Tk. 26 and costs
Tk. 17 to make, while each Model II sells for Tk. 19 and costs Tk. 14 to make. Each of the products must be processed
on two machines: Model I requires 3 hours on machine A and 2 hours on machine B, whereas Model II requires 7
hours on A and 1 hour on B. During the coming week machine A will be free for no more than 42 hours and machine
B for no more than 17 hours.
Requirements :
(i) Formulate the linear programming problem.
(ii) Solve the problem graphically and determine the number of each kind of products to be made in the coming week for
the company to maximize its profit.
(iii) What is the maximum profit ?
SOLUTION
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒎𝒆𝒏𝒕 ∶ (𝒊)
Let,
X = The number of units of Model – I
Y = The number of units of Model – II
∴ Profit Model – I = 26 – 17 = 9
∴ Profit Model – II = 19 – 14 = 5
∴ The objective function is 𝑍 = 9𝑥 + 5𝑦
The linear constraints are –
3𝑥 + 7𝑦 ≤ 42
2𝑥 + 𝑦 ≤ 17
( Objective function )
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 9𝑥 + 5𝑦
Subject t constraints ––
3𝑥 + 7𝑦 ≤ 42
2𝑥 + 𝑦 ≤ 17
And non-negativity constraints : (𝑥, 𝑦) ≥ 0
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒎𝒆𝒏𝒕 ∶ (𝒊𝒊)
We have,
3𝑥 + 7𝑦 = 42 –––––––––––––––––––– (𝑖)
2𝑥 + 𝑦 = 17 –––––––––––––––––––– (𝑖𝑖)
Putting x = 0 in (𝑖), we have
∴ 3 × 0 + 7𝑦 = 42
42
⟹ 𝑦 =
7
∴ 𝑦 = 6
Putting y = 0 in (𝑖), we have
∴ 3𝑥 + 7 × 0 = 42
42
⟹ 𝑥 =
3
∴ 𝑥 = 14
Hence, the points are (0,6) and (14,0).
Similarly, Putting x = 0 in (𝑖𝑖), we have
∴ 2 × 0 + 𝑦 = 17
∴ 𝑦 = 17
Putting y = 0 in (𝑖𝑖), we have
∴ 2𝑥 + 0 = 42
17
⟹ 𝑥 =
2
∴ 𝑥 = 8.5
Hence, the points are (0,17) and (8.5, 0).
Y-Values
18
16
14
12
10
8
6
4
2
0
0
2
4
6
8
10
12
14
16
Y-Values
From the graph, we have the following boundary point O(0,0), A(0,6), B(7,3), C(8.5,0)
Extreme Point
Co-ordinates (x,y)
Z = 9x+5y
A
B
C
(0, 6)
(7, 3)
(8.5, 0)
30
78
76.5
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒎𝒆𝒏𝒕 ∶ (𝒊𝒊𝒊)
Here the maximum profit Z = Tk. 76.5 at extreme point C(8.5,0) .
Previous – 02 : 21
A milkman has 70kgs of water and 130kgs of milk that it wishes to mix. A package of mixture I contains 300gms of
water and 700gms of milk, while mixture II contains 400gms of water and 600gms of milk per package. If a profit
of TK. 0.64 per package is obtained from mixture I and a profit of TK. 0.70 per package is obtained from II, how
many packages of each mixture should be made to obtain the maximum profit ?
SOLUTION
Let, X = The number of packages of Mixture – I
Y = The number of packages of Mixture – II
∴ Profit Model – I = 0.64
∴ Profit Model – II = 0.70
∴ The objective function is 𝑍 = 0. 64𝑥 + 0.70𝑦
The linear constraints are –
300𝑥 + 400𝑦 ≤ 70,000 ( 𝑔𝑚𝑠 )
700𝑥 + 600𝑦 ≤ 130,000 ( 𝑔𝑚𝑠 )
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 0.64𝑥 + 0.70𝑦
( Objective function )
Subject to constraints
300𝑥 + 400𝑦 ≤ 70,000
700𝑥 + 600𝑦 ≤ 130,000
And non-negativity constraints : (𝑥, 𝑦) ≥ 0
We have,
300𝑥 + 400𝑦 = 70,000 –––––––––––––––––––– (𝑖)
700𝑥 + 600 𝑦 = 130,000 –––––––––––––––––––– (𝑖𝑖)
Putting x = 0 in (𝑖), we have
∴ 300 × 0 + 400𝑦 = 70,000
70,000
⟹ 𝑦 =
400
∴ 𝑦 = 175
Putting y = 0 in (𝑖), we have
∴ 300𝑥 + 400 × 0 = 70,000
70,000
⟹ 𝑥 =
300
∴ 𝑥 = 233.33
Hence, the points are (0, 175) and (233.33, 0).
Similarly, Putting x = 0 in (𝑖𝑖), we have
∴ 700 × 0 + 600𝑦 = 130,000
130,000
⟹ 𝑦 =
600
∴ 𝑦 = 216.67
Putting y = 0 in (𝑖𝑖), we have
∴ 700𝑥 + 600 × 0 = 130,000
130,000
⟹ 𝑥 =
700
∴ 𝑥 = 185.71
Hence, the points are (0, 216.67) and (185.71, 0).
Now, we plot these points on a graph as follows :
Y-Values
250
200
150
100
50
0
0
50
100
150
200
250
Y-Values
From the graph, the following boundary points are O(0,0), A(0,175), B(100,100), C(185.71,0).
Extreme Point
Co-ordinates (x,y)
𝒁 = 𝟎. 𝟔𝟒𝒙 + 𝟎. 𝟕𝟎𝒚
O
A
B
C
(0, 0)
(0,175)
(100,100)
(185.71, 0)
0
122.50
134
118.86
Here the maximum value is Tk. 134 and occurs at (100,100). Hence, the milkman should make 100 packages mixture – I and 100
packages of mixture – II to obtain the minimum profit of Tk. 134. ( 𝑨𝑵𝑺 )
Previous – 3 : 19
XYZ steel company manufactures two types of product : Model E and Model D. Profit on two types of products are :
Tk. 9 and Tk. 7 to make one Model E is required 2 hours of machine 𝑀1, 1 hour of machine 𝑀2 and 4 hours of
machine M3. Similarly, to make Model D is required 1 hour of machine 𝑀1 , 2 hours of machine 𝑀2 and 5 hours of
machine 𝑀3 .
Production schedule indicates that during the coming week, 𝑀1 will be available for almost 30 hours, 𝑀2 for almost
24 hours and 𝑀3 for almost 72 hours.
Requirements :
(i) Formulate the linear programming problem.
(ii) Solve the problem graphically.
Solution
Let the decision variables be x and y.
Maximize 𝑍 = 9𝑥 + 7
( Objective function )
Subject to the constraints –
2𝑥 + 𝑦 ≤ 30 (𝑀1 )
𝑥 + 2𝑦 ≤ 24 (𝑀2 )
4𝑥 + 5𝑦 ≤ 72 (𝑀3 )
Non-negativity Constraint : x ≥ 0, y ≥ 0
Using equality, We have –
2𝑥 + 𝑦 = 30 –––––––––––––––––––– (𝑖 )
𝑥 + 2𝑦 = 24 –––––––––––––––––––– (𝑖𝑖)
4𝑥 + 5𝑦 = 72 –––––––––––––––––––– (𝑖𝑖𝑖)
Putting x = 0 in (𝑖), we have
∴ 2 × 0 + 𝑦 = 30
⟹ 𝑦 = 30
Putting y = 0 in (𝑖), we have
∴ 2𝑥 + 0 = 30
30
⟹ 𝑥 =
2
∴ 𝑥 = 15
Hence, the points are (0, 30) and (15, 0).
Similarly, Putting x = 0 in (𝑖𝑖), we have
∴ 0 + 2𝑦 = 24
∴ 𝑦 = 12
Putting y = 0 in (𝑖𝑖), we have
∴ 𝑥 + 2 × 0 = 24
⟹ 𝑥 = 24
Hence, the points are (0, 12) and (24, 0).
Putting x = 0 in (𝑖𝑖𝑖), we have
∴ 4 × 0 + 5𝑦 = 72
72
⟹ 𝑦 =
5
∴ 𝑦 = 14.4
Putting y = 0 in (𝑖𝑖𝑖), we have
∴ 4𝑥 + 5 × 0 = 72
72
⟹ 𝑥 =
4
∴ 𝑥 = 18
∴ The Points are (0, 14.4) and (18, 0)
Now, we plot these points on a graph as follows :
From the above graphic presentation, the boundary point of basic feasible solution – O(0, 0) ; A(0, 12) ; B(8, 8) ; C (13, 4)
and D(15, 0).
𝑬𝒙𝒕𝒓𝒆𝒎𝒆 𝑷𝒐𝒊𝒏𝒕𝒔
𝑪𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔(𝒙, 𝒚)
𝒁 = 𝟗𝒙 + 𝟕𝒚
O
(0, 0)
0
A
(0, 12)
84
B
(8, 8)
128
C
(13,4)
145
D
(15, 0)
135
∴ Hence, the maximum profit is Tk. 145, for finding this – x. = 13 and y = 4 Co-ordinate.
Previous – 4 : ( Class )
A diet of sick person must content at least 4,000 units of Vitamins,50 units of materials, 1,400 units of calcium. Foods A and
B are available at a cost Tk. 4 and Tk.3 per unit respectively. If 1 unit of A content 200 units of Vitamins, 1 unit of mineral
and 40 calories and 1 unit of B content 100 units of Vitamins, 2 units and 40 calories.
Fine that the graphical method, what combination of goods to have least cost.
SOLUTION
Decision Variables Food
Unit contents of
Cost ( TK. )
Vitamins Mineral Calories
𝑥
A
200
1
40
4
𝑦
B
100
2
40
3
Minimize 𝑍 = 4𝑥 + 3𝑦
( Objective function )
Subject to the constraints –
200𝑥 + 100𝑦 = 4,000
𝑥 + 2𝑦 = 50
40𝑥 + 40 𝑦 = 1,400
Non-negativity Constraint : x ≥ 0, y ≥ 0
Using equality, We have –
200x + 100y = 4,000 –––––––––––––––––––– ( 𝑖 )
𝑥 + 2𝑦 = 30 –––––––––––––––––––– ( 𝑖𝑖 )
4𝑥 + 4𝑦 =1,400 –––––––––––––––––––– ( 𝑖i𝑖 )
Putting x = 0 in (𝑖), we have
∴ 200 × 0 + 100𝑦 = 4,000
⟹ 𝑦 = 40
Putting y = 0 in (𝑖), we have
∴ 200𝑥 + 100 × 0 = 4,000
4,000
⟹ 𝑥 =
200
∴ 𝑥 = 20
Hence, the points are (0, 40) and (20, 0).
Similarly, Putting x = 0 in (𝑖𝑖), we have
∴ 0 + 2𝑦 = 50
∴ 𝑦 = 25
Putting y = 0 in (𝑖𝑖), we have
∴ 𝑥 + 2 × 0 = 50
⟹ 𝑥 = 50
Hence, the points are (0, 25) and (50, 0).
Putting x = 0 in (𝑖𝑖𝑖), we have
∴ 40 × 0 + 40𝑦 = 1,400
1,400
⟹ 𝑦 =
40
∴ 𝑦 = 35
Putting y = 0 in (𝑖𝑖𝑖), we have
∴ 40𝑥 + 40 × 0 = 1,400
1,400
⟹ 𝑥 =
40
∴ 𝑥 = 35
∴ The Points are (0, 35) and (35, 0)
Now, we plot these points on a graph as follows :
Y-Values
50
40
30
20
10
0
0
10
20
30
Y-Values
40
50
60
From the above graphic presentation, the boundary point of basic feasible solution – O(0, 0) ; A(0, 40) ; B(3, 8) ; C (20, 15)
and D(50, 0).
𝑬𝒙𝒕𝒓𝒆𝒎𝒆 𝑷𝒐𝒊𝒏𝒕𝒔
𝑪𝒐𝒐𝒓𝒅𝒊𝒏𝒂𝒕𝒆𝒔(𝒙, 𝒚)
𝒁 = 𝟒𝒙 + 𝟑𝒚
A
(0, 40)
120
B
(5, 3))
110
C
(20, 15)
125
D
(50, 0)
200
∴ Hence, the minimum cost is Tk. 110, for finding this – x. = 5 and y = 30 Co-ordinate.
PROBLEM – 5 : ( CLASS )
A firm produces two products X and Y respectively. The firm needs to produce X by 2 units labor and 1 unit capital.
Again to produce product needs 4 units of labor and 4 units of capital. The income of this firm from the product X
is Tk. 1 and from product Y is Tk. 3. The given labour amounts is 50 units and capital is 38 units.
SOlUTION
Requirement :
(i) Write the linear programming problem.
(ii) Determine what units are to be produced inputs maximize the income of the firm ?
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒎𝒆𝒏𝒕 ∶ (𝒊)
∴ The objective function is 𝑍 = 𝑥 + 3y
The linear constraints are –
2𝑥 + 4𝑦 ≤ 50
x + 4𝑦 ≤ 38
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍= 9𝑥 + 5𝑦
( Objective function )
Subject to constraints ––
2𝑥 + 4𝑦 ≤ 50
2𝑥 + 4𝑦 ≤ 38
And non-negativity constraints : 𝑥,𝑦 ≥ 0
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒎𝒆𝒏𝒕∶ (𝒊𝒊)
We have,
2x + 4y = 50 –––––––––––––––––––– (𝑖)
𝑥 + 4𝑦 = 38 –––––––––––––––––––– ( 𝑖𝑖 )
Putting x = 0 in (𝑖), we have
∴ 2 × 0 + 4𝑦 = 50
50
⟹𝑦= 4
∴ 𝑦 = 12.5
Putting y = 0 in (𝑖), we have
∴ 2𝑥 + 4 × 0 = 50
50
⟹𝑥= 2
∴ 𝑥 = 25
Hence, the points are (0, 12.5) and (25, 0).
Similarly, Putting x = 0 in (𝑖𝑖), we have
∴ 0 + 4y = 38
38
⟹ y= 4
∴ y = 9.5
Putting y = 0 in (𝑖𝑖), we have
∴ 𝑥 + 4 × 0 = 38
⟹ 𝑥 =38
Hence, the points are (0, 9.5) and (38, 0).
15
10
5
0
0
5
10
15
20
25
30
35
40
From the graph, we have the following boundary point O(0,0), A(0, 9.5), B(12, 6.5), C(25, 0).
Extreme Points
Co-ordinates (x, y)
Z = x + 3y
A
B
C
(0, 9.5)
(12, 6.5)
( 25, 0)
28.5
31.5
25
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒎𝒆𝒏𝒕 ∶ 𝒊𝒊𝒊
Here, the maximum profit is Tk. 31.5 at extreme point B(12, 6.5).
EXERCISE (18)
Safety lock company makes two kinds of locks : Model 'SS' (the super safe) and Model 'S' (the safe). Each model SS
lock sells for Tk. 24 and costs Tk.19 to make, while each model S lock sells for Tk. 18 and costs Tk. 15 to make.
Each of the locks must be processed on two machines. Model SS requires 3 hours on machine A and 2 hours on
machine B, whereas model S requires 7 hours on machine A and 1 hour on machine B. During the coming week
machine A will be free for no more than 42 hours and machine B for no more than 17 hours.
Requirements :
(i) Formulate the above linear programming problem.
(ii) Solve the problem graphically and determine the number of each kind of lock to be made in the coming week for
the company to maximize its profit.
(iii) What is the maximum profit ?