Example 2.1.A manufacturing company is producing two products A and B. Each of the products A and B requires the use of two machines P and Q. A requires 4 hours of processing on machine P and 3 hours of processing on machine Q. Product B requires 3 hours of processing on machine P and 6 hours of processing on machine Q. The unit profits for products A and Bare Rs 20 and Rs 30 respectively. The available time in a given quarter on machine P is 1000 hours and on machine Q is 1200 hours. The market survey has predicted 250 units of products A and 300 units of product B can be consumed in a quarter. The company is interested in deciding the product mix to maximize the profits. Formulate this problem as LP
model.
Sol. Formulating the problem in mathematical Equations
Let X_{A}= the quantity of product of type A manufactured in a quarter
X_{B}= the quantity of products of type B manufactured in a quarter
Z = the profit earned in a quarter.
(Objective function, which is to be maximized)
Therefore Z = 20 X_{A}+ 30 X_{B}
Z is to maximized under the following conditions:
4X_{A}+ 3X_{B ≤} 1000 (Time constraint of machine P)
3X_{A}+ 6X_{B} ≤ 1200 (Time constraint of machine Q)
X_{A ≤} 250 (Selling constraint of product A)
X_{B ≤} 300 (Selling constraint of product B)
X_{A}and X_{B} ≥0 (Condition of nonnegativity)
Example 2.2.MIS Steadfast Lid produces both interior and exterior house paints for wholesale distribution. Two types of raw materials A and B are used to manufacture the paints. The maximum availability of A is 10 tons a day and that of B is 15 tons a day daily requirement of raw material per ton of interior and exteriorpaints are as follow:
Requirement of Raw material Per ton of paint 

Raw material A 
3 
2 
10 
Raw material B 
2 
3 
15 
The market survey indicates that daily demand of interior paints cannot exceed that of exterior paints by 2 tons. The survey also shows that maximum demand of interior paints is only 3 tons daily.
The wholesale price per ton is Rs 75,000 for exterior paints and Rs 50,000 for interior paints.
Problem.How much interior and exterior paints MIS Steadfast should produce to maximize its profits?
Sol. Let X_{E} – tons of external paints to be produced daily.
X_{1} tons of internal paints to be produced daily.
Z – profit earned (objective function which is to be maximized)
Therefore
Z = 75000 X_{E}+ 50000 X_{1}
Z is to be maximized under the following constraints or conditions :
2X_{E}+ 3X_{1} ≤ 10 (Availability constraint of raw material A)
3X_{E} + 2X_{1} ≤ 15 (Availability constraint of raw material B)
X_{1} – X_{E} ≤ 2 (Demand constraint – Demand of interior paints daily cannot exceed more than 2 tons that of exterior paint)
X_{1} ≤ 3 (Demand of interior paint cannot exceed 3 tons every day)
Also,
X_{1} ≥ 0 (Nonnegativity constraint of interior paints)
X_{E ≥} 0 (Nonnegativity constraint of exterior paints)
The complete mathematical model for MIS Steadfast Ltd problem maybe written as follows :
Determine the tons of interior and exterior paints, X_{1} and. X_{E}to be produced in order to maximize Z = 75000 X_{E} + 50000 X, (objective function) under the constraints (conditions) of
2X_{E}+ 3X_{1} ≤ 10
3X_{E} + 2X_{1} ≤ 15
X_{1} X_{E} ≤ 2
X_{1} ≤ 3
X_{E} ≤ 0
X_{1}. ≥ 0
Let us verify if the above model satisfies the conditions of Linearity as we are using LP method to solve the problem, Linearity demands that Proportionality and Additivity must be satisfied.
1. Proportionality. It requires that usage of resources is directly proportional to that value of the variables, in this case X_{E}and X_{1}. Suppose MIS Steadfast have .a promotion policy that they will sell external paint at Rs 60,000 per ton after sales of exterior paintis more than 2 tons per day, then the equation of the objective function will no longer be true as each ton of external paint produced does not bring a revenue of Rs 75,000/ Actually for X_{E} ≤ 2 tons, the revenue will be Rs 75000 per ton and Rs 60,0001 per ton for X_{E} ≥ 2 tons. This situation does not satisfy the condition of direct proportionality with X_{E}.
2. Additivity demands that the objective function (Z) must be the direct sum of the individual contribution of different variables. For example, in case of two competing products like tea and coffee where an increase in sale of coffee adversely affects the sale of tea, these two products do not satisfy the conditions of additivity.
Since the mathematical model written above, satisfies both the properties of proportionality and additivity, we can use LP method to solve the problem.
Example 2.3.A chemical manufacturer produces two types of chemicals X and Y. Each typeof chemical is manufactured by a twostep process that involves machines A and B. The processing time for each unit of the two products 011 machines A and B are as follows :
Product 
Machine A (Hours/unit) 
Machine B (Hours/unit) 
X 
2 
3 
Y 
4 
2 
For a period of one month, the availability of machine A is 160 hours and availability of machine B is 120 hours. The manufacturer has found out from the market that maximum sale price of chemical X can be Rs 400 per unit and that of chemical Y can be Rs 500 per unit. Also, maximum of 20 units of chemical X can be sold in the market per month and 25 units of chemical B.
Problem.How much of chemical X and chemical Yshould be produced so that the profits can be maximized?
Sol. Let X – be the quantity of chemical X to be produced.
Y – be the quantity of chemical Y to be produced.
Z = 400X + 500Y
Z is to be maximized under the following constraints or conditions :
2X + 4 Y ≤ 160 (Availability constraint of machine A)
3X + 2Y ≤ 120 (Availability constraint of machine B)
X ≤20 (Marketing constraint of product X)
Y ≤ 25 (Marketing constraint of product B)
X ≥ 0 (Nonnegativity constraint of product X)
Y ≥ 0 (Nonnegativity constraint of product Y)
The complete mathematical model for the problem may be written as follows :
Determine the quantity of product X and Y to be produced in order to maximize
Z = 400X + 500Y (objective function) under the constraint of
2X +4Y ≤ 160
3X + 2Y ≤ 120
X ≤ 20
Y ≤ 25
X ≥ 0
Y ≥ 0
Example 2.4.MIS Plastics Ltd. produces two major products tables and chairs. Each table can be sold in the market at a profit of Rs 50 and chair can be sold at a profit of Rs 60. The requirement of man hours and machine hours for manufacturing of one table and one chair is as given below:
Product 
Manhours (Hours per unit) 
Machine (Hours per unit) 
Table 
10 
1.5 
Chair 
6 
1.0 
Total number of man hours available for this activity during the year is 1200 and machine is available only for 800 hours. The company carried out a survey to find out the maximum demand for their table and chairs. It was revealed that they can sell only 100 tables and 160 chairs in one year.
Problem.How many tables and chairs should MIS Plastics Ltd. produce to maximize their profit?
Sol. Let X be the number of tables to be produced.
Y be the number of chair to be produced.
Z = profit earned (objective function which is to be maximized)
Therefore Z = 50X + 60Y
Z is to be maximized under the following conditions
10X + 6Y ≤ 1200 (Availability constraint of man hours)
1· 5X + Y ≤ 800 (Availability constraint of machines hours)
X ≤ 100 (Demand constraint)
Y ≤ 160 (Demand constraint)
X ≥ 0 (Nonnegativity constraint of tables)
Y ≥ 0 (Nonnegativity constraint of chairs)
Example 2.5.An oil refinery uses blending process to produce gasoline in a typical manufacturing process. Crude A and B are mixed to produce gasoline G_{1}and gasoline G_{2}. The inputs and outputs of the process are as follows :
Availability of Crude A is only 200 tons and B 300 tons. Market demand of Gasoline G_{1}is 150 tons and Gasoline G_{2}is 120 tons. Profit by using process 1is Rs 200 per ton and by using process 2 is Rs 250 per ton. What is the optimal mix of two blending processes so that the refinery can maximize its profits?
Sol. Let X be the number of tons to be produced by process 1
Y be the number of tons to be produced by process 2
Z = Profit earned (Objective function which is to be maximized)
Therefore Z = 200X + 250Y
Z is to be maximized under the followingconditions
5X + 6Y ≤ 200
2X + 8Y ≤ 300
6X+ 5Y ≤ 150
8X + 7Y ≤ 120
X ≥ 0
Y ≥ 0
Example 2.6.Vitamin C and Vitamin Eare found in two different fruits F_{1} and F_{2}. One unit of fruit F_{1} Contains 3units of vitamin C and 2 units of vitamin E. Similarly, one unit of fruit F_{2} contains 2 units of vitamin C in it and 2 units of vitamin E in it. A patient needs minimum of 30 units of vitaminC and 20 units of vitamin E. Also one unit offruit F_{1} costs Rs 20 and oneunit of fruit F_{2} costs Rs 25. The problem, the hospitals faces is to find such units of ‘fruit FI and F!, which should be supplied to the patients at minimum cost.
Sol. Let X be the number of units of fruit F_{1}
Y be the number of unitsof fruit F_{2}
Z = Minimum cost (Objective function which is to be minimized)
Therefore Z = 20X + 25Y
Z is to be minimized subject to the following constraints :
3X + 2Y ≥ 30 (Minimum requirement of vitamin C)
2X + 2Y ≥ 20 (Minimum requirement of vitamin E)
X ≥ 0
Y≥ 0
Note – In this example a unit means different measures like grams and caloric values etc.
Example 2.7.Manufacturing company XYZ Lid manufactures two different types of products, refrigerators and washing machines. Boththese products have to be processed through two machines, Machine A and Machine B. Machine A is available for 200 hours and machine B is available for 100 hours. The requirement of time on these machines is as follows:
Refrigerator 
Washing machine 

Machine A 
10 
6 
Machine B 
5 
4 
The company makes a profit ofRs 800 on sale of one refrigerator and Rs 500 on sale of one washing machine. What quantities of refrigerators and washing machines should company XYZ Ltd. produce to maximize its profits?
Sol. Let X be the number of refrigerators to be manufactured.
and Y be the number of washing machines to be manufactured.
Z = Profit earned (Objective function which is to be maximized)
Therefore Z = 800X + 50Y
Z is to be maximized subject to the following constraints :
10X + 6y ≤ 200 (Availability of machine A)
5X + 4Y ≤ 100 (Availability of machine B)
X ≥ 0
Y ≥ 0
Example 2.8.A manufacturer can manufacture two different types of products, FRP sheets and FRP bath tubs. Each unit of FRP sheets of a particular size needs 5 kg of raw material A and 2 kg of raw material B. Each unit of FRP bath tubs needs 7 kg of raw material A and I kg of raw material B. Availability of raw material A in the market is 500 kg and that of raw material B 100 kg. Each FRP sheet contributes profit of Rs 100 and each FRP tub contributes profit of Rs 400. What is the most suitable product mix for the manufacturer to maximizeprofits?
Sol. Let X be the number of FRP sheets to be produced
and Y be the number of FRP tubs to be produced
Z = Profit earned (Objective function which is to be maximized)
Therefore Z = 100X + 400Y
Z is to be maximized subject to the following constraints
5X + 2Y ≤ 500
7X+ Y ≤ 100
X ≥ 0
Y ≥ 0
Example 2.9.A company manufactures three types of electrical products, electric iron, fan and toaster. All the three products have to be processed on two machines A and B. The processing time required by each product on both the machines is as given below:
Electric Iron 
Fan 
Toaster 

Machine A 
2 
3 
2 
Machine B 
1 
2 
3 
Machine A is available only for 200 hours and machine B is available for 160 hours. The firm should not manufacture more than 400 electric irons, more than 500 fails and more than 200 toasters. All electric iron gives a profit of Rs 110, a fall of Rs 150 and a toaster of Rs 80. What product mix would you recommend to the company so that its profits are maximized?
Sol. Let X_{1} be the number of electric irons to be manufactured.
X_{1}be the number of fans to be manufactured
X_{3} be the number of toasters to be manufactured
Z = profits generated (Objective functions which is to be maximized)
Z is to be maximized under the following constraints :
Z = 110X_{1}+ 150X_{2}+ 80X_{3}
2X_{1} + 3X_{2}+ 2X_{3} ≤ 200
X_{1} + 2X_{2} + 3X_{3 ≤} 160
Also, X_{1}, X_{2}, X_{3 ≥} 0
Example 2.10.A manufacturer of furniture makes twoproducts, chairs and tables. Processing of these products is done on two machines A and B. A chair requires two hours on machine A and 6 hours Oil machine B. A table requires 5 hours on machine A and notime onmachine B. There are 16 hours of time per day available on machine A and 30 hourson machine B. Profit gained by the manufacturer from a
chair and table is Rs 2 and Rs 10 respectively. What should be daily production of the two products?
Sol. Let X be the number of chairs to be manufactured
and Y be the number of tables to be manufactured
Let Z = Profit earned (objective function which is to be maximized)
Z is to be maximized under the following constraints
Z=2X+ 10Y
2X + 5Y ≤ 16 (Availability constraint of machine A)
6X ≤ 30 (Availability constraint of machine B)
X ≥ 0
Y ≥ 0
Example 2.11.Sandeep Electric company produces two products, motors and fans which are produced and sold onmonthly basis. The monthly production of motors cannot exceed 100 and that of fans 150 because of the limitation of the facilities of production. Motor requires 12 manhours of labour per month and fanrequires 12 man hours of labour per month. The company has a total of 50 employees. Profit margin on sale ofRs 110 and on sale of fan Rs 90. Formulate a LP problem.
Sol. Let X be the number of motors produced
and Y be the number of fans produced
Z = profit earned (objective function)
Z = 110X + 90Y
X ≤ 100 (Constraint of production of motors)
Y ≤ 150 (Constraint of production of fans)
30X + 12Y ≤ 50 (Constraint of availability of manhours)
X ≥ 0
Y ≥ 0
Example 2.12.A carpenter has 90, 80 and 50 running feet respectively of teak. plywood and rosewood. Product A requires 2, 1 and 1 running feet of teak, plywood and rosewood respectively. Product B requires 1, 2 and 1 running feet of teak, plywood and rosewood respectively. If A would sell/or Rs 48 and B would sell for Rs 40 per unit, how much of each should he make and sell in order to obtain the
maximum gross income out of his stock of wood ? Give a mathematical formulation to this linear programming problem.
Sol. Let X be the number of products A produced.
and Y be the number of products B produced.
Let Z = gross income (objective function)
Z is to be maximized under the following constraints
Z =48X +40Y
2X + Y ≤ 90 (Constraint of availability of teak)
X + 2Y ≤ 80 (Constraint of availability of plywood)
X + Y ≤ 50 (Constraint of availability of rosewood)
X ≥ 0
Y ≥ 0
Example 2.13.A retailer deals in two items only, item A and item B. He has Rs 50,000 to invest and a space to store at the most 60 pieces. An item of A costs himRs 2500 and B costs him 500. Net profit to him all items A is Rs 500 and on item B is Rs 150. If he can sell all the items that he purchases, how much should he invest his amount to have maximum profit? Give a mathematical formulation to the linear programming problem.
Sol. Let X be the number of items A the retailer should be purchase
and Y be the number of items B the retailer should purchase
Z = profit earned (objective function)
Z = 500X + 150Y
X + Y ≤ 60 (Constraint of total capacity)
2500X + 500 Y ≤ 50,000 (Constraint of total investment)
X ≥ 0
Y ≥ 0
Example 2.14.Company produces two types of products A and B. Product B is superior quality and product B is lower quality. Profit all the two types of products are Rs 30 and Rs 40 respectively. The data about resources required and availability of resources are given below:
Requirement of Resources 

Product A 
Product B 
Capacity available (per month) 

Raw material (Kg) 
60 
120 
12000 
Machine hours (per piece) 
8 
5 
630 
Assembly 
3 
4 
500 
How should the company manufacture the two types of products in order to get the maximum overall profits? Formulate the problem as a LPP.
Sol. Let X be the quantity of product A manufactured
and Y be the quantity of product B manufactured.
Z = overall profits (Objective function)
Zis to be maximized under the following conditions
Z = 30X + 40Y
60x + 120Y ≤ 12000 (Constraint ofraw material)
8X + 5Y ≤ 630 (Constraint of machine hours)
3X + 4Y ≤500 (Constraint of assembly)
X ≥0
Y ≥0
Example 2.15.A machine is producing either product A or B. Itcan produce product A by using 2 units of chemicals and 1unit of a compound and can produce B by using 1unit of chemical and 2 units of compound. Only 800 units of chemical and 100 units of compound are available. The profits available per unit of A and Bare Rs 30 and Rs 20 respectively. Find the optimum allocation of units between A and B tomaximize the total profit. Give a mathematical formulation to the LP problem.
Sol. Let X be the quantity of product A produced
andY be the quantity of product B produced
Z = Total profits (Objective function)
Z is to be maximized under the following constraints.
Z = 30X + 20Y
2X + Y ≤ 800 (Constraint of availability of chemical)
X + 2Y ≤1000 (Constraint of availability of compound)
X≥0
Y≥0
Example 2.16.ABC Ltd manufacturers two products A and B. For manufacturing product A, a machine has tobe used for 3 hours and the operator has toput in4 hours. For manufacturing product B the machine use is for 4 hours and the operator works only for 2 hours. In one week. the company can use amaximum of 60 hours of machine time and the worker is available for 80 hours. The contribution of profit by product A is Rs 40 and by product B is Rs 50. Formulate the problem as a LP model andfind what quantities should ABC Lid manufacture of product A and B tobe able tomaximize their profits. Assume that all quantities of A and B produced are sold without any problem.
Sol. Let X and Y be the quantities of product A and B respectively to be manufactured, The data of the problem can be put in the form of a table as shown below.
Decision Variable 
Product 
Time Required 
Product/Unit 

Machine 
Operator 

X 
A 
2 
4 
40 
Y 
B 
4 
2 
50 
Max time available (Hours) 
60 
78 
Objective function
Maximise Z = 40X + 50Y
Constraints
(i) 2X + 4Y ≤ 60
(ii) 4X + 2Y ≤ 78 Converting the inequalities into equalities for solving these
(iii) X, Y ≥ 0
Now X and Y can be solved to have maximum of Z.
X= 16
Y=7
Z=40 ×16+50 ×7=640+350=Rs990
ABC should manufacturer 16 number of component A and 7 of component B.
Example 2.17.An ancillary unit of HMT tractors produces two types of tractor parts A and B. All the parts that are manufactured by the ancillary units are purchased by HMT. The manufacturing cost of part A is Rs 2400 and of part B it is Rs 3200. HMT purchases part A for Rs 3200 and part Bfor Rs 3600. The company’s present production capacity is limited because of the following three constraints.
(a) Budget constraints – cash available at the beginning of year Rs 10,50,000
(b) Machine time available 4200 hours/year
(c) Manufactures available/year 5600.
Part A needs 8 hours of machine time and part B needs only 4 hours. Also, the manhours requiredfor part A are 8 hours and for part B 8 hours, find out what quantities of these should be produced by the ancillary unit to maximize its profits,
Sol. The problem can be formulated as an LP problem with constraints as follows. The given data can be put in the form of a table for ease of understanding let X_{1}be the quantity of part A to be manufactured and X, be the quantity of part B to be manufactured
Decision Variable Quantities to be manufactured 
Part 
Machine Time (Hours) 
Operator Time (Hours) 
Cost Price Rs 
Selling Price (Rs) 
X_{1} 
A 
8 
20 
2400 
3200 
X_{2} 
B 
4 
8 
3200 
3600 
Total availability 
4200 
5600 

Budget availability 
=Rs 25,60,000 
Objective Function
Maximize Z = 800X_{1}+ 400X_{2}
2400 X_{1} + 3200 X_{2}≤4,60,000 (Constraints of Budget)
8X_{1}+ 4X_{2}≤ 4200 (Constraint of machine time)
20X_{I} + 8X_{2}≤5600 (Constraint of man hours)
X_{1}, X_{2}≥ 0 (nonnegativity constraints)
These inequalities can be converted into equations and solved.
3200 X_{1}+ 3600 X_{2}= 25,60,000
8X_{1} + 4X_{2}= 4200
90X_{1}+ 8X_{2}= 5600
X_{1} = 200
X_{2}= 650
It can be seen that the cost of quantity of parts manufactured remains within the total budget.
Also Z = Rs 4,20,000.
Example 2.18.A company supplying three types of parts to an automatic manufacturing company, purchases castings of three parts from a nearby foundry and performs three types of operators before selling these and cost per hour of these machines is given in the table below
Machine 
Capacity/hour 
Capacity/hour 
Capacity/hour 
Cost/hour (Rs.) 
Cutting 
20 
60 
25 
150 
Drilling 
40 
20 
40 
100 
Polishing 
50 
50 
20 
200 
The cost ofthe castings for A, Rs 120, for B Rs 200, for C Rs 400 and the selling price of these parts is Rs 200, Rs 350 and Rs 500 respectively. All the parts that are processed by the company can be sold, What quantity of various parts should the company process for selling in order to maximize their profits?
Sol. Let X_{1}, X_{2} and X_{3} be the number of parts the company should process.
Example 2.19.ABC Lid is assembling two products P_{1} and P_{2}, The cost of assembling one unit of products P_{1} and P_{2} is Rs 200 and Rs 240 respectively. The availability of work station for two products is limited to 60 hours and the two products spend 6 hours and 2 hours respectively all tile work station. The products can be sold for Rs 280 and Rs 320 respectively. Total mallhours available are 400 and P_{1} requires 2 manhours and P_{2} requires 4 manhours, Formulate the problem as a LPP.
Sol. Let X_{1}and X_{2}be the number of products P_{1}and P_{2}to be assembled.
Maximise Z = 80X_{1}+ 80X_{2} with the following constraints
6X_{1}+ 2X_{2}≤ 60 (Workstation constraints)
2X_{1} + 4X_{2}≤400 (manhours constraints)
X_{1},X_{2}≥0
These inequalities can be converted into equations and solved for X_{1}and X_{2}.
Example 2.20.A person has 12,00,000 which he wants to investin different types of investment opportunities. He has consulted investment consultant to advise him on this who has given him the following advice.
(a) Government sector, PoswfficeiNSC etc upto 40%
(b) Share X and Share Y both put together Rs 2,00.000.
(c) Mutual funds A and B. At least 25%
The rate of return which he gets from these investments are
(a) Government sectors 0 ·1
(b) Share X 0·2
(c) Share Y0.25
(d) Mutual fund A 0 ·15
(e) Mutual fund B 0 ·12
Formulate the above as a LP problem.
Sol. LetX_{1}= Government sector investment
X_{2}= Share X
X_{3}= Share Y
X_{4}= Mutual Fund A
X_{5}= Mutual Fund B
Objective function is to be maximized = Z = 0.1 X, + 0.2 X_{2} + 0.25 X_{3} + 0.15 X4 + 0.12 X_{5} the following constraints.
X_{1} + X_{2} + X_{3}+ X_{4}+ X_{5}≤ 12,00,000 (total investment constraint)
X≤4,80,000
X_{2}+ X_{3}≤ 2,00,000
X_{4} + X_{5}≥3,00,000
X_{1}, X_{2}, X_{3}, X_{4}, X_{5}≥ 0
These inequalities can be converted into equations and solved for X_{1}, X_{2} , X_{3} , X_{4} and X_{5}and hence for objective function Z.
Example 2.21.A firm manufacturing three products A, B and C. Time to manufacture product A is twice that for B and thrice that for C and they are to be produced in the ratio 2: 3 : 4. The relevant data is given below. If the whole labour is engaged in manufacturing product A. 2000 units of the product call be produced. There is a demand for at least 200. 300 and 400 units of the product A. B. C and the profit earned per unit is Rs 100, Rs 70 and Rs 50 respectively. Formulate the problem as a linear programming problem.
Raw Material 
Requirement per unit of product 
Total availability (Kg) 

A 
B 
C 

P 
6 
5 
9 
4000 
Q 
4 
8 
6 
5000 
Maximize Z=100A+70B+50C (Objective function) subject to the following constraints
6A + 58 + 9C≤ 4000 (Constraint of raw material P)
4A + 88 +6C ≤ 5000 (Constraint of raw material Q)
Example 2.22.A patient wants to decide the constituents of diet which his daily
requirement of proteins, fats and carbohydrate’s at the minimum cost. The choice is to be made from three different types offoods. The yields per unit of these foods are as follows:
Food Type 
Yield per unit 
Cost per unit (Rs.) 

Proteins  Fats  Carbohydrates  
1 
2 
2 
4 
40 
2 
4 
2 
4 
50 
3 
8 
10 
8 
60 
Minimum requirement 
1200 
400 
800 
Formulate linear programming model for the problem.
Sol. Let A, Band C be the number of units of food type 1,2,3, and respectively.
Maximise Z = 40A + SOB + 60C (Objective function) subject to the following constraints)
2A + 4B + 8C ≥ 1200 (Constraint of Proteins)
2A + 2B + 10C ≥400 (Constraints of Fats)
4A + 4B + 8C ≥ 800 (Constraints of Carbohydrates)
A≥0
B ≥ 0 (Nonnegativity constraints)
C≥0
Example 2.23.A firm manufactures three products A, Band C. The profits are Rs3. Rs2 and Rs 4 respectively. The firm has two machines and the required processing time in minutes for each machine on each product if given below:
Product 

A 
B 
C 

Machine 
C 
4 
3 
5 
D 
2 
2 
4 
Machine C and D have 2000 and 2500 machineminutes respectively. The firm must manufacture 100 A’s, 200 B’s and 50 C’s but 110t more than 150 A’s, Set lip all LP model to maximize the profit.
Sol. Maximize Z: 3A + 2B + 4C
Subject to 4A + 38 + 5C ≥ 2000
2A + 28 + 4C ≥ 2500
A ≥ 100
B ≥ 200
A ≤ 150
C ≤ 50
A, B, C ≥ 0 .
Example 2.24.A [inn can produce three types of cloth A, Band C. Three kinds of wool are required for it. say, red wool. green and blue wool. One unit length of type A cloth needs 2 yards of red wool and 3 yards of blue wool, one unit length of type B cloth needs 3 yards of red wool. 2 yards of green wool and 2 yards of blue wool and one unit length of type C cloth needs 5 yards of green wool and 4 yards of blue wool. The firm has a stock of only 8 yards of red wool. 10 yards of green wool and 15 yards of blue wool. It is assumed that the income obtained from one unit of type A cloth is Rs3. of type B cloth is Rs5 and that of type C cloth is Rs4. Formulate the problem as linear programming problem,
Sol. Maximize Z: 3A + 58 + 4C (Objective function) subject to the following constraints
2A + 38 ≤ 8 (Constraints of availability of red wool)
2A + 5B ≤ 10 (Constraints of availability of green wool)
3A + 28 + 4C ≤ 15 (Constraints of availability of blue wool)
A, 8, C ≥ 0 (Nonnegativity constraints)
Example 2.25.Orient Paper Mills produces two grades of paper X and Y. Because of raw material restrictions not more than 400 tons of grade X and no/more than 300 tons of grade Y can be produce a ton of week. There are 160 production hours in a week. It requires 0·2 and 0.4 hours to produce a ton of products X and Y respectively. with corresponding profit of Rs 20 and Rs 50 per ton. Formulate a linear programming problem to optimize the product mix for maximum profit.
Sol. Maximize Z = 20X + 50Y (Objective function)
Subject to the constraints
0·2X + 0·4Y ≤ 160 (Constraints of production hours)
X ≤ 400 (Constraints of raw material)
Y ≤ 300 (Constraints of raw material)
X ≥ 0
Y ≥ 0 (Nonnegative constraints)
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