In the graph, the feasible region for a LPP is shaded. The objective function Z = 2x

Engineering

Mathematics

maximum and minimum of quadratic and rational expression

Linear Programming

Linear Inequalities

Question

In the graph, the feasible region for a LPP is shaded. The objective function Z = 2x – 3y, will be minimum at :

(4, 10)

(6, 8)

(0, 8)

(6, 5)

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To minimize Z = 2x - 3y, we evaluate it at all corner points of the feasible region. The point with the smallest Z value is the solution.

Corner points from the graph: (0,8), (4,10), (6,8), (6,5).

Calculate Z for each:

At (0,8): Z = $2 \times 0 - 3 \times 8 = -24$

At (4,10): Z = $2 \times 4 - 3 \times 10 = 8 - 30 = -22$

At (6,8): Z = $2 \times 6 - 3 \times 8 = 12 - 24 = -12$

At (6,5): Z = $2 \times 6 - 3 \times 5 = 12 - 15 = -3$

The minimum value of Z is -24 at (0,8).

Final Answer: (0, 8)