Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
The system of constraints is:
x + 2y ≤ 120 ...(i)
x + y ≥ 60 ....(ii)
x - 2y ≥ 0 ....(iii)
and x, y ≥ 0 ....(iv)
Let l1 : x + 2y = 120
l2 : x + y = 60
l3 : x - 2y = 0
It is observed that the feasible region CADE is bounded.
The coordinates of C, A, D, E are (60, 0), (120, 0), (60, 30), (40, 20) respectively.
Thus, we use the Corner Point Method to determine the maximum and minimum values of Z.
We have : Z = 5x + l0y
| Corner Point | Corresponding values of Z |
| (60, 0) | 300 (Minimum) |
| (120, 0) | 600 |
| (60, 30) | 600 |
| (40, 20) | 400 |
Hence Zmin = 300 at (60, 0) and Zmax = 600 at all points on the line segment joining the points (120, 0) and (60, 30).
APPEARS IN
संबंधित प्रश्न
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:
Maximise Z = 3x + 2y
subject to x + 2y ≤ 10, 3x + y ≤ 15, 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 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.
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 = 3x + 4y if the feasible region (shaded) for a LPP is shown in Figure
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.
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
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 corner points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y ______.
Compare the quantity in Column A and Column B
| Column A | Column B |
| Maximum of Z | 325 |
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 = ______.
In a LPP, the linear inequalities or restrictions on the variables are called ____________.
The feasible region for an LPP is always a ______ polygon.
Maximum value of the objective function Z = ax + by in a LPP always occurs at only one corner point of the feasible region.
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.
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:
Objective function of a linear programming problem is ____________.
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. If M and m respectively be the largest and smallest values at corner points then ____________.
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.
Maximize Z = 10 x1 + 25 x2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
Z = 6x + 21 y, subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
Maximize Z = 10×1 + 25×2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
