Question

The feasible region of an LPP is shown in the figure. If Z = 11x + 7y then the maximum value of Z occurs at

• A. (0, 5)
• B. (3, 3)
• C. (3, 2)
• D. (5, 0)

Answer 1

The Correct Option is D

We are given a linear programming problem (LPP) with the objective function $ Z = 11x + 7y $, and the feasible region defined by the constraints $ x + y = 5 $ and $ x + 3y = 9 $.

Step 1: Find the points of intersection of the lines.

The equations of the lines are:

1. $ x + y = 5 $
2. $ x + 3y = 9 $

We solve these two equations to find the points of intersection.

Solve $ x + y = 5 $ for $ x $:

$$ x = 5 - y $$

Substitute this into the second equation $ x + 3y = 9 $:

$$ (5 - y) + 3y = 9 $$

Simplify:

$$ 5 + 2y = 9 $$ $$ 2y = 4 \quad \Rightarrow \quad y = 2 $$

Substitute $ y = 2 $ back into $ x + y = 5 $:

$$ x + 2 = 5 \quad \Rightarrow \quad x = 3 $$

Thus, the point of intersection is $ (3, 2) $.

Step 2: Check the values of $ Z $ at the vertices of the feasible region.

The vertices of the feasible region are $ (0, 5) $, $ (5, 0) $, and $ (3, 2) $. Now, calculate $ Z = 11x + 7y $ at each vertex:

• At $ (0, 5) $: $$ Z = 11(0) + 7(5) = 35 $$
• At $ (5, 0) $: $$ Z = 11(5) + 7(0) = 55 $$
• At $ (3, 2) $: $$ Z = 11(3) + 7(2) = 33 + 14 = 47 $$

The maximum value of $ Z $ occurs at $ (5, 0) $ with $ Z = 55 $.

Answer 2

Intersection points of the lines are:

$$ (x+3y) - (x+y) = 9 - 5 \implies 2y = 4 \implies y = 2. $$

Substitute $ y = 2 $ into $ x+y=5 $:

$$ x + 2 = 5 \implies x = 3. $$

The intersection point is $ (3, 2) $. The corner points of the feasible region are $ (0, 3) $, $ (3, 2) $, and $ (5, 0) $. Now calculate $ Z = 11x + 7y $ at each corner point:

• $ Z(0, 3) = 11(0) + 7(3) = 21 $
• $ Z(3, 2) = 11(3) + 7(2) = 33 + 14 = 47 $
• $ Z(5, 0) = 11(5) + 7(0) = 55 $

The maximum value of $ Z $ is 55, which occurs at $ (5, 0) $.

Final Answer: The final answer is $ {(5, 0)} $.