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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.
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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.
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