Question

In L.P.P., the maximum value of objective function $Z = 6x + 3y$ subject to $x + y \leq 5, x + 2y \geq 4, 4x + y \leq 12, x, y \geq 0$ is \dots

• A. 132/7
• B. 22
• C. 15
• D. 122/7

Hint:

Always substitute a calculated intersection point back into the remaining un-used inequalities to verify it actually lies inside the feasible region before trusting its $Z$ value!

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The maximum value occurs at one of the corner points of the feasible region defined by the linear inequalities.

Step 2: Formula Application:
Solve pairs of equations to find intersection points: 1. $x+y=5$ and $4x+y=12 \implies 3x=7 \implies x=7/3, y=8/3$. 2. $x+2y=4$ and Y-axis $\implies (0, 2)$. 3. $x+y=5$ and X-axis $\implies (5, 0)$.

Step 3: Explanation:
Test corner points in $Z = 6x + 3y$: - At $(0, 2): Z = 6(0) + 3(2) = 6$. - At $(0, 5): Z = 15$. - At $(3, 0): Z = 18$. - At $(7/3, 8/3): Z = 6(7/3) + 3(8/3) = 14 + 8 = 22$. Wait, checking $4x+y \leq 12$ for $(7/3, 8/3)$: $4(2.33) + 2.66 = 9.33 + 2.66 = 12$. Valid. Checking $(7/3, 8/3)$ in $x+y \leq 5$: $2.33 + 2.66 = 5$. Valid. The maximum value calculated is 22.

Step 4: Final Answer:
The maximum value is 22 (Option B).