Formulation of Linear Programming Problem - Advance Level

Formulation of Linear Programming Problem - Advance Level

Dear Readers,

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. Mohan wants to invest the total amount of Rs. 15,000 in saving certificates and national saving bonds. According to rules, he has to invest at least Rs. 2000 in saving certificates and Rs. 2500 in national saving bonds. The interest rate is 8% on saving certificate and 10% on national savingbonds per annum. He invests Rs. x in saving certificates and Rs. y in national saving bonds. Then the objective function for this problem is

•  0.08x + 0.10y
•  (x/2000) + (y/2500)
•  2000x + 2500y
•  (x/8) + (y / 10)

Solution
The objective function for this problem is 0.08x + 0.10y

Q2.Two tailors A and B earn Rs. 15 and Rs. 20 per day respectively A can make 6 shirts and 4 pants in a day while B can make 10 shirts and 3 pants. To spend minimum on 60 shirts and 40 pants, A and B work x and y days respectively. Then linear constraints except x>= 0,y >= 0, are

•  15x+20y >= 0 , 60x + 40y >= 0
•  15x + 20y >= 0, 6x + 10y = 10
•  6x + 10y >= 60, 4x + 3y >= 40
•  6x + 10y <= 60, 4x + 3y <= 40

Solution
Linear constraints except x>= 0,y >= 0, are 6x + 10y <= 60, 4x + 3y <= 40

Q3.  In the examination of P.E.T. the total marks of mathematics are 300. If the answer is right, marks provided is 3 and if the answer is wrong, marks provided is –1. A student knows the correct answer of 67 questions and remaining questions are doubtful for him. He takes the time 1 ¹ / ₂ minute to give the correct answer and 3 minute that for doubtful. Total time is 3 hour. In the question paper after every two simple questions, one question is doubtful. He solves the questions one by one, then the number of questions solved by him, is

•  67
•  90
•  79
•  80

Solution
The number of questions solved by the student, is 90

Q4. A shopkeeper wants to purchase two articles A and B of cost price Rs. 4 and Rs. 3 respectively. He thought that he may earn 30 paise by selling article A and 10 paise by selling article B He has not to purchase total articles of more than Rs. 24. If he purchases the number of articles of A and B, x and y respectively, then linear constraints are

•  x>=0,y>=0,4x+3y <= 24
•  x>=0,y>=0,30x + 10y <= 24
•  x>=0,y>=0,4x+3y>=24x
•  x>=0,y>=0,30x + 40y >= 24

Solution
Linear constraints are x>=0,y>=0,4x+3y <= 24

Q5.A company manufacturers two types of products A and B. The storage capacity of its godown is 100 units. Total investment amount is Rs. 30,000. The cost prices of A and B are Rs. 400 and Rs. 900 respectively. All the products are sold and per unit profit is Rs. 100 and Rs. 120 through A and B respectively. If x units of A and y units of B be produced, then two linear constraints and iso-profit line are respectively

•  x+y = 100 , 4x + 9y = 300, 100x + 120y = x
•  x+y <= 100, 4x+9y <= 300, x + 2y = c
•  x+y <= 100, 4x + 9y <= 300, 100x + 120y = c
•  x+y >= 100, 9x + 4y >= 300, 5x + 6y = c

Solution
 Two linear constraints and iso-profit line are respectively x+y <= 100, 4x + 9y <= 300, 100x + 120y = c

Q6. We have to purchase two articles A and B of cost Rs. 45 and Rs. 25 respectively. I can purchase total article maximum of Rs. 1000. After selling the articles A and B, the profit per unit is Rs. 5 and 3 respectively. If I purchase x and y numbers of articles A and B respectively, then the mathematical formulation of problem is

•  x >= 0, y >= 0,45x + 25y >= 1000, 5x + 3y = c
•  x>=0,y>=0,45x+25y <= 1000, 5x+3y = c
• x>= 0,y >=0,45x+25y <= 1000, 3x+5y = c
•  None of these

Solution
The mathematical formulation is x>=0,y>=0,45x+25y <= 1000, 5x+3y = c