CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Linear Programming

Q.5. A manufacturer produces two types of steel trunks. He has two machines, A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type requires 3 hours on machine A and two hours on machine B. Machines A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of Rs.30 per trunk on the first type of trunk and Rs.25 per trunk on the second type. Formulate a Linear Programming Problem to find out how many trunks of each type he must make each day to maximize his profit.

Solution :

Fig.

Let x unit of type A and y unit of type B be produced each day.
Therefore , z = 30 x + 25 y
Under the restrictions :
3 x + 3 y ≤ 18
3 x + 2 y ≤ 15
x ≥ 0 , y ≥ 0 .
Corner points of shaded portion are : (0,0),(0,6),(5,0),(3,3).
Maximum z is at (3,3) .
Therefore , x = 3 and y = 3 . [Ans.]

Q.6. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs50 per kg to purchase Food I and Rs70 per kg to purchase Food II. Formulate this problem as a linear programming problem to minimize the cost of such a mixture.

Solution :

Let the mixture contain x kg of Food I and y kg of Food II. Then x ≥ 0, y ≥ 0.

Since the mixture must contain at least 8 units of vitamin A and 10 units of vitamin C, then we have the constraints as :
2x + y ≥ 8 and x + 2y ≥ 10.
Total cost Z of purchasing x kg of food I and y kg of food II is given by
Z = 50 x + 70 y
Hence , the mathematical formulation of the problem is :
Minimize Z = 50 x + 70 y ---------------------------- (1)
Subject to the constraints :
2x + y ≥ 8 ------------------------------- (2)
x + 2y ≥ 0 ------------------------------- (3)
x, y ≥ 0 ------------------------------- (4)

Fig

from the graph , the feasible region is unbounded. Let us evaluate Z at the corner points A(0,8), B(2,4) and C(10,0).

Hence, the minimum value of Z is Rs380. [Ans.]

Paper By Mr. M.P.Keshari
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