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 Question:
We have a Linear Programming Problem (L.P.P.) with an objective function $Z = 6x + 3y$ to maximize, subject to three linear constraints and non-negativity constraints in the first quadrant.

Step 2: Key Formula or Approach:
1. Find the corner points (vertices) of the bounded feasible region by solving the equations of the intersecting bounding lines.
2. Evaluate the objective function $Z$ at each valid corner point.
3. The maximum value of $Z$ will occur at one of these vertices.

Step 3: Detailed Explanation:
Let's analyze the lines and find their points of intersection:
$L_1$: $x + y = 5$
$L_2$: $x + 2y = 4$
$L_3$: $4x + y = 12$
Intersection of $L_1$ and $L_3$:
From $L_1$, $y = 5 - x$. Substitute into $L_3$:
$4x + (5 - x) = 12 \implies 3x = 7 \implies x = 7/3$.
$y = 5 - 7/3 = 8/3$.
Point $A(7/3, 8/3)$. Is it valid? Check $L_2$: $7/3 + 2(8/3) = 23/3 \ge 4$. Yes, it is in the feasible region.
Evaluate $Z$ at $A$: $Z_A = 6(7/3) + 3(8/3) = 14 + 8 = 22$.

Intersection of $L_2$ and $L_3$:
From $L_2$, $x = 4 - 2y$. Substitute into $L_3$:
$4(4 - 2y) + y = 12 \implies 16 - 8y + y = 12 \implies 7y = 4 \implies y = 4/7$.
$x = 4 - 2(4/7) = 20/7$.
Point $B(20/7, 4/7)$. Is it valid? Check $L_1$: $20/7 + 4/7 = 24/7 \le 5$. Yes.
Evaluate $Z$ at $B$: $Z_B = 6(20/7) + 3(4/7) = 120/7 + 12/7 = 132/7 \approx 18.85$.

Now check the Y-axis intercepts ($x=0$):
$L_1 \implies (0,5)$ (Max limit on Y due to $\le$). Valid! $Z = 3(5) = 15$.
$L_2 \implies (0,2)$ (Min limit on Y due to $\ge$). Valid! $Z = 3(2) = 6$.

Check X-axis intercepts ($y=0$):
$L_3 \implies (3,0)$ (Max limit on X). Check $L_2$: $3 + 2(0) = 3$. But $L_2$ requires $\ge 4$. So $(3,0)$ is OUTSIDE the feasible region.
There are no valid corner points on the X-axis for this specific bounded shape.
Comparing the valid values: 22, 18.85, 15, 6.
The absolute maximum is 22.

Step 4: Final Answer:
The maximum value is 22, matching option (b).