Advertisements
Advertisements
Question
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
Solution
The shaded region is the feasible region determined by the constraints x ≤ 3, y ≤ 2, x ≥ 0, y ≥.
The feasible region is bounded with four corners O(0, 0), A(3, 0), B(3, 2) and C(0, 2).
So, the maximum value can occur at any corner
Let us evaluate the value of Z.
| Corner points | Value of Z | |
| O(0, 0) | 11(0) + 7(0) = 0 | |
| A(3, 0) | 11(3) + 7(0) = 33 | |
| B(3, 2) | 11(3) + 7(2) = 47 | ← Maximum |
| C(0, 2) | 11(0) + 7(2) = 14 |
Hence, the maximum value of the function Z is 47 at (3, 2).
APPEARS IN
RELATED QUESTIONS
Solve the following Linear Programming Problems graphically:
Minimise Z = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Solve the following Linear Programming Problems graphically:
Minimise Z = x + 2y
subject to 2x + y ≥ 3, x + 2y ≥ 6, x, 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?
A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin content of one kg food is given below:
| Food | Vitamin A | Vitamin B | Vitamin C |
| X | 1 | 2 | 3 |
| Y | 2 | 2 | 1 |
One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet?
A manufacturer makes two types of toys A and B. Three machines are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:
| Type of toy | Machines | ||
| I | II | III | |
| A | 12 | 18 | 6 |
| B | 6 | 0 | 9 |
Each machine is available for a maximum of 6 hours per day. If the profit on each toy of type A is Rs 7.50 and that on each toy of type B is Rs 5, show that 15 toys of type A and 30 of type B should be manufactured in a day to get maximum profit.
A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is Rs 100 and that on a bracelet is Rs 300. Formulate on L.P.P. for finding how many of each should be produced daily to maximize the profit?
It is being given that at least one of each must be produced.
The feasible region for a LPP is shown in figure. Evaluate Z = 4x + y at each of the corner points of this region. Find the minimum value of Z, if it exists.
Refer to quastion 12. What will be the minimum cost?
Refer to question 13. Solve the linear programming problem and determine the maximum profit to the manufacturer
Refer to question 15. Determine the maximum distance that the man can travel.
Maximise Z = x + y subject to x + 4y ≤ 8, 2x + 3y ≤ 12, 3x + y ≤ 9, x ≥ 0, y ≥ 0.
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.
Refer to Question 27. Maximum of Z occurs at ______.
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 32, Maximum of F – Minimum of F = ______.
In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same ______ value.
A feasible region of a system of linear inequalities is said to be ______ if it can be enclosed within a circle.
The feasible region for an LPP is always a ______ polygon.
In the given graph, the feasible region for an LPP is shaded. The objective function Z = 2x – 3y will be minimum at:
Objective function of a linear programming problem is ____________.
In linear programming infeasible solutions
In Corner point method for solving a linear programming problem, one finds the feasible region of the linear programming problem, determines its corner points, and evaluates the objective function Z = ax + by at each corner point. Let M and m respectively be the largest and smallest values at corner points. In case feasible region is unbounded, M is the maximum value of the objective function if ____________.
In Corner point method for solving a linear programming problem, one finds the feasible region of the linear programming problem, determines its corner points, and evaluates the objective function Z = ax + by at each corner point. Let M and m respectively be the largest and smallest values at corner points. In case the feasible region is unbounded, m is the minimum value of the objective function.
Maximize Z = 3x + 5y, subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Maximize Z = 6x + 4y, subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, x ≥ 0, y ≥ 0.
