Linear Programming
Let the merchant stock x desktop computers and y portable computers.
Let P be the profit.
Table
Type |
Number |
Cost per computer (Rs.) |
Investment (Rs.) |
Profit (Rs.) |
Desktop |
x |
25000 |
25000 x |
4500 x |
Portable |
y |
40000 |
40000 y |
5000 y |
Total |
x + y |
25000 x + 40000 y |
4500 x + 5000 y |
We are to maximise
P = 4500 x + 5000 y
subject to constraints
x + y ≤ 250
25000 x + 40000 y ≤ 7000000
or 5 x + 8 y ≤ 1400
x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies a ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 250
For x = 0, y = 250
For y = 0, x = 250
∴ line meets OX in A(250, 0) and OY in L(0, 250)
Again we draw the graph of 5 x + 8 y = 1400
For x = 0, 8 y = 1400 or y = 175
For y = 0, 5 x = 1400 or y = 280
∴ line meets OX in B(280, 0) and OY in M(0, 175).
Since feasible region satisfies all the constraints.
∴ OACM is the feasible region.
The comer points are O(0, 0), A(250, 0), C(200, 50), M(0, 175)
At O(0, 0), P = 4500 × 0 + 5000 × 0 = 0 + 0 = 0
At A(250, 0), P = 4500 × 250 + 5000 × 0 = 1125000
At C(200, 50), P = 4500 × 200 + 5000 × 50 = 900000 + 250000 = 1150000
At M(0, 175), P = 4500 × 0 + 5000 × 175 = 0 + 875000 = 875000
∴ maximum value = 1150000 at (200, 50)
∴ 200 units of desktop models and 50 units of portable models are to be stocked for maximum profit of Rs. 1150000.
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Minimise and Maximise Z = 3x + 9y
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x ≤ y
x ≥ 0, y ≥ 0
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