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Question
Minimise z = – 3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 What will be the minimum value of z ?
Options
– 12
12
6
– 6
Solution
– 12
Explanation:
Objective function z = – 3x + 4y
Constraints are x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0
(i) Consider the line, x + 2y = 8
It pass through A(8, 0) and B(0, 4)
Putting x = 0, y = 0 in x + 2y ≤ 8, 0 ≤ 8 which is true
⇒ Region x + 2y ≤ 8 lies on and below AB
(ii) The line 3x + 2y = 12 passes through P(4, 0), Q(0, 6)
Putting x = 0, y = 0 in 3x + 2y ≤ 12
⇒ 0 ≤ 12, which is true
∴ Region 3x + 2y ≤ 12 lies on and below PQ.
(iii) x ≥ 0, the region lies on and to the right of y-axis.
(iv) y ≥ 0 lies on and above x-axis.
(v) Solving the equations x + 2y = 8 and 3x + 2y = 12
We get x = 2, y = 3 R is (2, 3) where AB and PQ intersect.
The shaded region OPRB is the feasible I region.
At P(4, 0) | z = – 3x + 4y = – 12 + 0 = – 12 |
At R(2, 3) | z = – 6 + 12 = 6 |
At B(0, 4) | z = 0 + 16 = 16 |
At Q(0, 0) | z = 0 |
Thus minimum value of z is – 12 at P(4, 0)