Advertisements
Advertisements
Question
Maximise Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0
Advertisements
Solution
Given that: Z = 3x + 4y and the constraints x + y ≤ 1, x ≥ 0, y ≥ 0
Let x + y = 1
| x | 1 | 0 |
| y | 0 | 1 |
The shaded area OAB is the feasible region determined by x + y ≤ 1, x ≥ 0, y ≥ 0
The feasible region is bounded.
So, maximum value will occur at the corner points O(0, 0), A(1, 0), B(0, 1).
Now, evaluating the value of Z, we get
| Corner points | Value of Z | |
| O(0, 0) | 3(0) + 4(0) = 0 | |
| A(1, 0) | 3(1) + 4(0) = 3 | |
| B(0, 1) | 3(0) + 4(1) = 4 | ← Maximum |
Hence, the maximum value of Z is 4 at (0, 1).
APPEARS IN
RELATED QUESTIONS
Solve the following Linear Programming Problems graphically:
Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Show that the minimum of Z occurs at more than two points.
Minimise and Maximise Z = x + 2y
subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200; 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 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.
To maintain his health a person must fulfil certain minimum daily requirements for several kinds of nutrients. Assuming that there are only three kinds of nutrients-calcium, protein and calories and the person's diet consists of only two food items, I and II, whose price and nutrient contents are shown in the table below:
| Food I (per lb) |
Food II (per lb) |
Minimum daily requirement for the nutrient |
||||
| Calcium | 10 | 5 | 20 | |||
| Protein | 5 | 4 | 20 | |||
| Calories | 2 | 6 | 13 | |||
| Price (Rs) | 60 | 100 |
What combination of two food items will satisfy the daily requirement and entail the least cost? Formulate this as a LPP.
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
Feasible region (shaded) for a LPP is shown in Figure. Maximise Z = 5x + 7y.
The feasible region for a LPP is shown in Figure. Find the minimum value of Z = 11x + 7y
In figure, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y.
Refer to question 14. How many sweaters of each type should the company make in a day to get a maximum profit? What is the maximum profit.
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.
In order to supplement daily diet, a person wishes to take some X and some wishes Y tablets. The contents of iron, calcium and vitamins in X and Y (in milligrams per tablet) are given as below:
| Tablets | Iron | Calcium | Vitamin |
| X | 6 | 3 | 2 |
| Y | 2 | 3 | 4 |
The person needs atleast 18 milligrams of iron, 21 milligrams of calcium and 16 milligrams of vitamin. The price of each tablet of X and Y is Rs 2 and Rs 1 respectively. How many tablets of each should the person take in order to satisfy the above requirement at the minimum cost?
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 solution for a LPP is shown in Figure. Let Z = 3x – 4y be the objective function. Minimum of Z occurs at ______.
Refer to Question 27. Maximum of Z occurs at ______.
Refer to Question 27. (Maximum value of Z + Minimum value of Z) is equal to ______.
A corner point of a feasible region is a point in the region which is the ______ of two boundary lines.
In a LPP, the maximum value of the objective function Z = ax + by is always finite.
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 one that is concerned with ____________.
In linear programming infeasible solutions
In linear programming, optimal solution ____________.
A maximum or a minimum may not exist for a linear programming problem 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 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 = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Maximize Z = 10×1 + 25×2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
Which condition represents non-negative restrictions for two decision variables?
Which statement defines an Optimisation Problem?
Which other constraint must be satisfied together with \[5x+y\leq 100\] in the profit-maximization formulation?
What is the name of the quantity to be optimised in linear programming?
