Linear Programming

We are to minimise Z = 3x + 2y subject to the constraints x + y ≥ 8 , 3x + 5 y ≤ 15, x ≥ 0, y ≥ 0.
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of x + y = 8
For x = 0, y = 8
For y = 0, x = 8
∴ line meets OX in A(8, 0) and OY in L(0, 8).
Again we draw the graph of 3x + 5y = 15

For x = 0, 5y = 15, or y = 3
For y = 0, 3x = 15 or x = 5
∴ line meets OX in B(5, 0) and OY in M(0, 3).
Since feasible region satisfies all the constraints.
∴ in this case, feasible region is empty and hence no feasible solution.

We are to maximise
Z = x + y subject to the constraints x - y ≤ - 1, - x + y ≤ 0, x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of
x - y = - 1
For x = 0, - y = - 1 ⇒ y = 1
For y = 0, x = - 1
∴  line x - y = 1 meets OX in A (- 1, 0) and OY in B (0, 1)
Again we draw the graph of - x + y = 0
For x = 0, y = 0
For y = 0, x = 0
∴ line - x + y = 0 passes through O (0, 0).

Since feasible region satisfies all the constraints
∴  in this case, feasible region is empty
∴ there exists no solution to the given linear programming problem.

Let x be the number of tables and y be the number of chairs.
Let Z be the profit
∴    We are to maximize
Z = 50x + 15y subject to the constraints
250x + 50y ≤ 5000
x + y ≤ 60
x, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Let us draw the graph of 250 x + 50y = 5000
For x = 0, 50 y = 5000 ⇒ y = 100
For y = 0, 250 x = 5000 ⇒ x = 20
∴ line 250 x + 50 y = 5000 meets OX in A (20, 0) and OY in B (0, 100)
Again we draw the graph of the line x + y = 60
For x = 0, y = 60
For y = 0, x = 60
∴ line x + y = 60 meets OX in C (60, 0) and OY in D (0, 60).
Since feasible is the region which satisfies all the constraints
∴ feasible region is the quadrilateral OAED. The comer points are O (0, 0). A (20, 0), E (10, 50), D (0, 60)

At O (0, 0), Z = 0 + 0 = 0
At A (20, 0), Z = 50 (20) + 15 (0) = 1000 + 0 = 1000
At E(10, 50), Z = 50 (10) + 15 (50) = 500 + 750= 1250
At D(0, 60), Z = 50 (0) + 15 (60) = 0 + 900 = 900
∴  maximum value of Z = 1250 at E (10, 50)
∴  maximum profit = Rs. 1250
when x = 10, y = 50 i.e., when number of tables = 10, number of chairs = 50

Let x hectare of land be allocated to crop X and y hectare to crop Y. Clearly, x ≥ 0, y ≥ 0.
Profit per hectare on crop X = Rs. 10500
Profit per hectare on crop Y = Rs. 9000
∴  total profit = Rs. (10500 x + 9000 y)
The mathematical formulation of the problem is as follows:
Maximise    Z = 10500 x + 9000 y
subject to constraints x + y ≤ 50, 20x + 10y ≤ 800
i.e. 2 x + y ≤ 80
and    x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of x + y = 50
For x = 0, y = 50
For y = 0, x = 50
∴ line meets OX in A(50, 0) and OY in L(0, 50).
Again we draw the graph of 2x + y = 80.
For x = 0, y = 80
For y = 0, 2x = 80 or x = 40

∴  line meets OX in B(40, 0) and OY in M(0, 80).
Since feasible region satisfies all the constraints.
∴ OBCL is the feasible region.
The corner points are O(0, 0), B(40, 0), C(30, 20), L(0, 50).
At O(0, 0), Z = 0 + 0 = 0
At B(40, 0), Z = 420000 + 0 = 420000
At C(30, 20), Z = 315000 + 180000 = 495000
At L(0, 50), Z = 0 + 450000 = 450000
∴ maximum value = 495000 at (30, 20)
∴ society will get the maximum profit of Rs. 495000 by allocating 30 hectares for crop X and 20 hectares for crop Y.