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Minimise z = – 3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0 What will be the minimum value of z ? -

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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

MCQ

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)

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