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प्रश्न
If the feasible region for a LPP is ______ then the optimal value of the objective function Z = ax + by may or may not exist.
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उत्तर
If the feasible region for a LPP is open unbounded then the optimal value of the objective function Z = ax + by may or may not exist.
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संबंधित प्रश्न
Solve the following Linear Programming Problems graphically:
Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, y ≥ 0.
Solve the following Linear Programming Problems graphically:
Minimise Z = – 3x + 4 y
subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
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 = 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.
Minimise and Maximise Z = 5x + 10 y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, 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.
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?
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?
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 Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
Minimise Z = 13x – 15y subject to the constraints: x + y ≤ 7, 2x – 3y + 6 ≥ 0, x ≥ 0, y ≥ 0
Determine the maximum value of Z = 3x + 4y if the feasible region (shaded) for a LPP is shown in Figure
The feasible region for a LPP is shown in Figure. Find the minimum value of Z = 11x + 7y
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.
Refer to Question 27. (Maximum value of Z + Minimum value of Z) is equal to ______.
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.
If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax + by may or may not exist.
Maximum value of the objective function Z = ax + by in a LPP always occurs at only one corner point of the feasible region.
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:
Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
A linear programming problem is one that is concerned with ____________.
In linear programming, optimal solution ____________.
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.
In a LPP, the objective function is always ____________.
Maximize Z = 7x + 11y, subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Maximize Z = 6x + 4y, subject to x ≤ 2, x + y ≤ 3, -2x + y ≤ 1, 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.
