Advertisements
Advertisements
प्रश्न
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?
Advertisements
उत्तर
Let the airline sell x tickets of executive class and y tickets of economy class.
The mathematical formulation of the given problem is as follows.
Maximize z = 1000x + 600y … (1)
subject to the constraints,

The feasible region determined by the constraints is as follows.
The corner points of the feasible region are A (20, 80), B (40, 160), and
C (20, 180).
The values of z at these corner points are as follows.
| Corner point | z = 1000x + 600 | |
| A (20, 80) | 68000 | |
| B (40, 160) | 136000 | → Maximum |
| C (20, 180) | 128000 |
The maximum value of z is 136000 at (40, 160).
Thus, 40 tickets of executive class and 160 tickets of economy class should be sold to maximize the profit and the maximum profit is Rs 136000.
APPEARS IN
संबंधित प्रश्न
Two tailors, A and B, earn Rs 300 and Rs 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP
Solve the following Linear Programming Problems graphically:
Maximise Z = 5x + 3y
subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0
Solve the following Linear Programming Problems graphically:
Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Show that the minimum of Z occurs at more than two points.
Maximise Z = – x + 2y, Subject to the constraints:
x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.
Show that the minimum of Z occurs at more than two points.
Maximise Z = x + y, subject to x – y ≤ –1, –x + y ≤ 0, x, y ≥ 0.
A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag?
The minimum value of the objective function Z = ax + by in a linear programming problem always occurs at only one corner point of the feasible region
Determine the maximum value of Z = 11x + 7y subject to the constraints : 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
Refer to question 15. Determine the maximum distance that the man can travel.
A manufacturer produces two Models of bikes-Model X and Model Y. Model X takes a 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.
A company makes 3 model of calculators: A, B and C at factory I and factory II. The company has orders for at least 6400 calculators of model A, 4000 calculator of model B and 4800 calculator of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made every day; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made everyday. It costs Rs 12000 and Rs 15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.
The feasible region for an LPP is shown in the figure. Let F = 3x – 4y be the objective function. Maximum value of F is ______.
Refer to Question 30. Minimum value of F is ______.
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5). Let F = 4x + 6y be the objective function. The Minimum value of F occurs at ______.
Refer to Question 32, Maximum of F – Minimum of F = ______.
In a LPP, the linear inequalities or restrictions on the variables are called ____________.
In a LPP, the objective function is always ______.
If the feasible region for a LPP is ______ then the optimal value of the objective function Z = ax + by may or may not exist.
A corner point of a feasible region is a point in the region which is the ______ of two boundary lines.
Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function Z = 3x + 9y maximum?

A linear programming problem is as follows:
Minimize Z = 30x + 50y
Subject to the constraints: 3x + 5y ≥ 15, 2x + 3y ≤ 18, x ≥ 0, y ≥ 0
In the feasible region, the minimum value of Z occurs at:
For an objective function Z = ax + by, where a, b > 0; the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20), (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum Z occurs at both the points (30, 30) and (0, 40) is:
In a linear programming problem, the constraints on the decision variables x and y are x − 3y ≥ 0, y ≥ 0, 0 ≤ x ≤ 3. The feasible region:
Objective function of a linear programming problem is ____________.
Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
In linear programming infeasible solutions
In linear programming, optimal solution ____________.
Maximize Z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
Maximize Z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Maximize Z = 10 x1 + 25 x2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
Maximize Z = 10×1 + 25×2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
The feasible region for an LPP is shown shaded in the following figure. Minimum of Z = 4x + 3y occurs at the point.

