Consider the linear programmig problem:

Minimize z = 50x + 70y

Subject to 2x + y ≥ 8, x + 2y ≥ 10, x ≥ 0, y ≥ 0

The minimum value of objective function is:

380

We are to minimise z = 50x + 70y

subject to the constraints

2x + y ≥ 8

x + 2y ≥ 10

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 2x + y = 8

For x = 0, y = 8

For y = 0, 2 x = 8 or x = 4

∴ line meets OX in A (4, 0) and OY in L (0, 8).

Again we draw the graph of x+ 2y= 10.

For x = 0, 2y = 10 or y = 5

For y= 0, x= 10

∴ line meets OX in B (10, 0) and OY in M (0, 5)

Since feasible region is the region which satisfies all the constraints

∴ shaded region is the feasible region and corner points are B (10, 0), C (2, 4), L (0, 8).

At B(10, 0), z= 50(10)+ 70(0)= 500+0 = 500

At C (2, 4), z = 50 (2) +70 (4) = 100 + 280 = 380

At L (0, 8), z = 50 (0) + 70 (8) = 0 + 560 = 560

∴ 38 is the smallest value of z at (2, 4)

Since feasible region is unbounded

∴ we are to check whether this value is minimum.

For this we draw the graph of

50x + 70y < 380 ...(1)

Since (1) has no common point with feasible region. 

∴ minimum value = 380 at (2, 4).