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:
Answer & explanation
Correct answer: option 1
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).