Important Questions: Linear Programming | Mathematics (Maths) Class 12 - JEE PDF Download

Q1: Solve the following linear programming problem graphically:
Minimise Z = 200 x + 500 y subject to the constraints:
x + 2y ≥ 10
3x + 4y ≤ 24
x ≥ 0, y ≥ 0
Ans: Given,
Minimise Z = 200 x + 500 y … (1)
subject to the constraints:
x + 2y ≥ 10 … (2)
3x + 4y ≤ 24 … (3)
x ≥ 0, y ≥ 0 … (4)
Let us draw the graph of x + 2y = 10 and 3x + 4y = 24 as below.
The shaded region in the above figure is the feasible region ABC determined by the system of constraints (2) to (4), which is bounded. The coordinates of corner point A, B and C are (0,5), (4,3) and (0,6) respectively.
Calculation of Z = 200x + 500y at these points.
Hence, the minimum value of Z is 2300 is at the point (4, 3).

Q2: A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 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 Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem to minimise the cost of such a mixture.
Ans: Let the mixture contain x kg of Food ‘I’ and y kg of Food ‘II’.
Clearly, x ≥ 0, y ≥ 0.
Tabulate the given data as below.
Given that, the mixture must contain at least 8 units of vitamin A and 10 units of vitamin C.
Thus, the constraints are:
2x + y ≥ 8
x + 2y ≥ 10
Total cost Z of purchasing x kg of food ‘I’ and y kg of Food ‘II’ is Z = 50x + 70y
Hence, the mathematical formulation of the problem is:
Minimise Z = 50x + 70y … (1)
subject to the constraints:
2x + y ≥ 8 … (2)
x + 2y ≥ 10 … (3)
x, y ≥ 0 … (4)
Let us draw the graph of 2x + y = 8 and x + 2y = 10 as given below.
Here, observe that the feasible region is unbounded.
Let us evaluate the value of Z at the corner points A(0,8), B(2,4) and C(10,0).
Therefore, the minimum value of Z is 380 obtained at the point (2, 4).
Hence, the optimal mixing strategy for the dietician would be to mix 2 kg of Food ‘I’ and 4 kg of Food ‘II’, and with this strategy, the minimum cost of the mixture will be Rs 380.

Q3: Solve the following LPP graphically:
Maximise Z = 2x + 3y, subject to x + y ≤ 4, x ≥ 0, y ≥ 0
Ans: Let us draw the graph pf x + y =4 as below.
The shaded region (OAB) in the above figure is the feasible region determined by the system of constraints x ≥ 0, y ≥ 0 and x + y ≤ 4.
The feasible region OAB is bounded and the maximum value will occur at a corner point of the feasible region.
Corner Points are O(0, 0), A (4, 0) and B (0, 4).
Evaluate Z at each of these corner points.
Hence, the maximum value of Z is 12 at the point (0, 4).

Q4: A manufacturing company makes two types of television sets; one is black and white and the other is colour. The company has resources to make at most 300 sets a week. It takes Rs 1800 to make a black and white set and Rs 2700 to make a coloured set. The company can spend not more than Rs 648000 a week to make television sets. If it makes a profit of Rs 510 per black and white set and Rs 675 per coloured set, how many sets of each type should be produced so that the company has a maximum profit? Formulate this problem as a LPP given that the objective is to maximise the profit.
Ans: Let x and y denote, respectively, the number of black and white sets and coloured sets made each week.
Thus x ≥ 0, y ≥ 0
The company can make at most 300 sets a week, therefore, x + y ≤ 300.
Weekly cost (in Rs) of manufacturing the set is 1800x + 2700y and the company can spend up to Rs. 648000.
Therefore, 1800x + 2700y ≤ 648000
or
2x + 3y ≤ 720
The total profit on x black and white sets and y coloured sets is Rs (510x + 675y).
Let the objective function be Z = 510x + 675y.
Therefore, the mathematical formulation of the problem is as follows.
Maximise Z = 510x + 675y subject to the constraints :
x + y ≤ 300
2x + 3y ≤ 720
x ≥ 0, y ≥ 0
The graph of x + y = 30 and 2x + 3y = 720 is given below.