Advertisements
Advertisements
Question
In figure, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y.
Advertisements
Solution
Here, corner points are given as follows:
`"R"(7/2, 3/4)`
`"Q"(3/2, 15/4)`
`"P"(3/13, 24/13)`
And `"S"(18/7, 2/7)`
Now, evaluating the value of Z for the feasible region RQPS.
| Corner points | Value of Z = x + 2y | |
| `"R"(7/2, 3/4)` | Z = `7/2 + 2(3/4) = 5` | ← Maximum |
| `"Q"(3/2, 15/4)` | Z = `3/2 + 2(15/4) = 9` | |
| `"P"(3/13, 24/13)` | Z = `3/13 + 2(24/13) = 51/13` | |
| `"S"(18/7, 2/7)` | Z = `18/7 + 2(2/7) = 22/7` | ← Minimum |
Hence, the maximum value of Z is 9 at `(3/2, 15/4)` and the minimum value of Z is `22/7` at `(18/7, 2/7)`
APPEARS IN
RELATED QUESTIONS
Solve the following Linear Programming Problems graphically:
Maximise Z = 3x + 4y
subject to the constraints : x + y ≤ 4, x ≥ 0, 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.
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 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.
If the feasible region for a linear programming problem is bounded, then the objective function Z = ax + by has both a maximum and a minimum value on R.
Maximise Z = 3x + 4y, subject to the constraints: x + y ≤ 1, 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. Evaluate Z = 4x + y at each of the corner points of this region. Find the minimum value of Z, if it exists.
A man rides his motorcycle at the speed of 50 km/hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 80 km/hour, the petrol cost increases to Rs 3 per km. He has atmost Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem
Refer to question 13. Solve the linear programming problem and determine the maximum profit to the manufacturer
Refer to Question 27. Maximum of Z occurs at ______.
Refer to Question 27. (Maximum value of Z + Minimum value of Z) is equal to ______.
The feasible region for an LPP is shown in the figure. Let F = 3x – 4y be the objective function. Maximum 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 = ______.
A feasible region of a system of linear inequalities is said to be ______ if it can be enclosed within a circle.
In a LPP, the minimum value of the objective function Z = ax + by is always 0 if the origin is one of the corner point of the feasible region.
In the given graph, the feasible region for an LPP is shaded. The objective function Z = 2x – 3y will be minimum at:

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:
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:
A linear programming problem is one that is concerned with ____________.
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.
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 = 10×1 + 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.

