Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
Two tailors, A and B, earn Rs 300 and Rs 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP
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 = 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.
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.
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 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.
An aeroplane can carry a maximum of 200 passengers. A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?
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.
Determine the maximum value of Z = 11x + 7y subject to the constraints : 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.
Maximise the function Z = 11x + 7y, subject to the constraints: x ≤ 3, y ≤ 2, x ≥ 0, y ≥ 0.
The feasible region for a LPP is shown in Figure. Find the minimum value of Z = 11x + 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 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.
Refer to question 15. Determine the maximum distance that the man can travel.
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 27. Maximum of Z occurs at ______.
Refer to Question 27. (Maximum value of Z + Minimum value of Z) is equal to ______.
If the feasible region for a LPP is ______ then the optimal value of the objective function Z = ax + by may or may not exist.
If the feasible region for a LPP is unbounded, maximum or minimum of the objective function Z = ax + by may or may not exist.
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.
Based on the given shaded region as the feasible region in the graph, at which point(s) is the objective function Z = 3x + 9y maximum?
In the given graph, the feasible region for an LPP is shaded. The objective function Z = 2x – 3y will be minimum 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 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 = 10 x1 + 25 x2, subject to 0 ≤ x1 ≤ 3, 0 ≤ x2 ≤ 3, x1 + x2 ≤ 5.
