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.
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.
Refer to Example 9. How many packets of each food should be used to maximize the amount of vitamin A in the diet? What is the maximum amount of vitamin A in the diet?
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?
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
Minimise Z = 13x – 15y subject to the constraints: x + y ≤ 7, 2x – 3y + 6 ≥ 0, 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
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 13. Solve the linear programming problem and determine the maximum profit to the manufacturer
Refer to question 15. Determine the maximum distance that the man can travel.
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.
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 |
Refer to Question 32, Maximum of F – Minimum of F = ______.
In a LPP, the linear inequalities or restrictions on the variables are called ____________.
In a LPP, the objective function is always ______.
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 a LPP, the maximum value of the objective function Z = ax + by is always finite.
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 ____________.
Z = 7x + y, subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at ____________.
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. 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 ____________.
If two corner points of the feasible region are both optimal solutions of the same type, i.e., both produce the same maximum or minimum.
Maximize Z = 3x + 5y, subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
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.
