Formulation of Linear Programming Problem - Basic Level

Linear programming is a method to achieve the best outcome in a mathematical model whose requirements are represented by linear relationships. Linear programming is a special case of mathematical programming. .

Q1. A whole sale merchant wants to start the business of cereal with Rs. 24,000. Wheat is Rs. 400 per quintal and rice is Rs. 600 per quintal. He has capacity to store 200 quintal cereal. He earns the profit Rs. 25 per quintal on wheat and Rs. 40 per quintal on rice. If he store x quintal rice and y quintal wheat, then for maximum profit the objective function is
•  25x + 40y
•  40x + 25y
•  400x + 600y
•  (400x/40) + (600y /25)
Solution
maximum profit the objective function is 40x+25y

Q2.A firm produces two types of product A and B. The profit on both is Rs. 2 per item. Every product need processing on machines M1 and M2. For A, machines M1 and M2 takes 1 minute and 2 minute respectively and that of for B, machines M1 and M2 takes the time 1 minute and 1 minute. The machines M1, M2 are not available more than 8 hours and 10 hours any of day respectively. If the products made x of A and y of B, then the linear constraints for the L.P.P. except are
•  x + y <= 480, 2x + y <= 600
•  x+y<=8, 2x+y <= 10
•  x+y >= 480, 2x+y >= 600
•  x+y<=8 ,2x+y >= 10
Solution
Then the linear constraints for the L.P.P. are x + y <= 480, 2x + y <= 600

Q3.  In a test of Mathematics, there are two types of questions to be answered, short answered and long answered. The relevant data are given below
 Time taken to solve Marks Number of questions Short answered questions 5 minutes 3 10 Long answered questions 10 minutes 5 14
The total marks are 100. Student can solve all the questions. To secure maximum marks, student solve x short answered and y long answered questions in three hours, then the linear constraints except x>>=0,y>=0 are
•  5x+10y <= 180, x<= 10,y<= 14
•  x+10y >= 180, x<=10,y<=14
•  5x+10y >= 180, x>=10,y>=14
•  5x+10y <= 180, x>=10,y>=14
Solution
the linear constraints except x>>=0,y>=0 are 5x+10y <= 180, x<= 10,y<= 14

Q4. A company manufactures two types of telephone sets A and B. The A type telephone set requires 2 hour and B type telephone requires 4 hour to make. The company has 800 work hour per day. 300 telephone can pack in a day. The selling prices of A and B type telephones are Rs. 300 and 400 respectively. For maximum profit company produces x telephones of A type and y telephones of B type. Then except x>=0 and y>=0, linear constraints are
•  x + 2y <= 400, x+y <= 300 maxz = 300x + 400y
•  2x+y<=400, x+y >= 300 maxz = 400x + 300y
•  2x+y >= 400, x+y >= 300 maxz = 300x+400y
•  x+2y <= 400, x+y >= 300 maxz = 300x+400y
Solution
Linear constraints are x + 2y <= 400, x+y <= 300 maxz = 300x + 400y

Q5.In a factory which produces two products A and B, in manufacturing product A, the machine and the carpenter requires 3 hours each and in manufacturing product B, the machine and carpenter requires 5 hour and 3 hour respectively. The machine and carpenter works at most 80 hour and 50 hour per week respectively. The profit on A and B are Rs. 6 and Rs. 8 respectively.If profit is maximum by manufacturing x and y units of A and B type products respectively, then for the function 6x+8y, the constraints are
•  x>=0, y>= 0, 5x+3y <= 80,3x + 2y <= 50
•  x>=0,y>=0,3x+3y<=50
•  x>=0,y>=0,3x+5y >= 80,2x+3y >= 50
•  x>=0,y>=0,5x+3y>=80,3x+2y>=50
Solution
for the function 6x+8y, the constraints are x>=0,y>=0,3x+3y<=50

Q6. The sum of two positive integers is at most 5. The difference between two times of second number and first number is at most 4. If the first number is x and second number y, then for maximizing the product of these two numbers, the mathematical formulation is
•  x+y>=5,2y-x>=4,x>=0,y>=0
•  x+y>=5,-2x+y>=4,x>=0,y>=0
• x+y<=5,2y-x<=4,x>=0,y>=0
•  None of these
Solution
the mathematical formulation is x+y<=5,2y-x<=4,x>=0,y>=0

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Formulation of Linear Programming Problem - Basic Level
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