Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Let factory I be operated for x days and II for y days.
At factory I: 50 calculators of model A and at factory II, 40 calculators of model A are made everyday.
Company has orders of atleast 6400 calculators of model A.
∴ 50x + 40y ≥ 6400
⇒ 5x + 4y ≥ 640
Also, at factory I, 50 calculators of model B and at factory II, 20 calculators of model B are made everyday
Company has the orders of atleast 4000 of calculators of model B.
∴ 50x + 20y ≥ 4000
⇒ 5x + 2y ≥ 4000
Similarly for model C,
30x + 40y ≥ 4800
⇒ 3x + 4y ≥ 480
And x ≥ 0, y ≥ 0
It costs ₹ 12,000 and ₹ 15000 to operate the factories I and II each day.
∴ Required LPP is
Minimise Z = 12000x + 15000y subject to the constraints
5x + 4y ≥ 640 ......(i)
5x + 2y ≥ 400 .......(ii)
3x + 4y ≥ 480 .......(iii)
x ≥ 0, y ≥ 0 .......(iv)
Table for (i) equation 5x + 4y = 640
| x | 0 | 128 |
| y | 160 | 0 |
Table for (ii) equation 5x + 2y = 400
| x | 0 | 80 |
| y | 200 | 0 |
Table for (iii) equation 3x + 4y = 480
| x | 0 | 160 |
| y | 120 | 0 |
On solving equation (i) and (iii), we get
x = 80, y = 60
On solving equation (i) and (ii) we get
x = 32 and y = 120
From the graph, we see that the feasible region ABCD is open unbounded whose corners are A(160, 0), B(80, 60), C(32, 120) and D(0, 200).
Let us find the values of Z.
| Corner points | Value of Z = 12000x + 15000y | |
| A(160, 0) | Z = 12000(160) + 0 = 1920000 | |
| B(80, 60) | Z = 12000(80) + 15000(60) = 1860000 | ← Minimum |
| C(32, 120) | Z = 12000(32) + 15000(120) = 2184000 | |
| D(0, 200) | Z = 0 + 15000(200) = 3000000 |
From the above table, it is clear that the value of Z = 1860000 may or may not be minimum for an open unbounded region.
Now, to decide this, we draw a graph of 12000x + 15000y < 1860000
⇒ 4x + 5y < 620
And we have to check whether there is a common point in this feasible region or not.
So, from the graph, there is no common point.
∴ Z = 12000x + 15000y has minimum value 1860000 at (80, 60).
Factory I: 80 days
Factory II: 60 days.
APPEARS IN
संबंधित प्रश्न
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 = 3x + 5y
such that x + 3y ≥ 3, x + y ≥ 2, 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 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.
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.
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
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
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 quastion 12. What will be the minimum cost?
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.
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.
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?
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 ______.
A feasible region of a system of linear inequalities is said to be ______ if it can be enclosed within a circle.
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 ____________.
The maximum value of the object function Z = 5x + 10 y subject to the constraints x + 2y ≤ 120, x + y ≥ 60, x - 2y ≥ 0, x ≥ 0, y ≥ 0 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. If M and m respectively be the largest and smallest values at corner points then ____________.
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 = 10×1 + 25×2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
