For the objective function $Z=4x+6y$ subject to the constraints $3x + 2y ≥ 5,7x + 2y ≤9,x ≥ 0, y ≥0$, the maximum value of Z occurs at (a, b) and the minimum value of Z occurs at (p, q) then the value of $\frac{a}{p}+\frac{b}{q}$ is:

4.5

The correct answer is Option (2) → 4.5

The feasible region is formed by the constraints:

• $3x + 2y \geq 5$
• $7x + 2y \leq 9$
• $x \geq 0, y \geq 0$

First, find the corner points of the feasible region:

1. Intersection of $3x + 2y = 5$ and $7x + 2y = 9$

Subtracting gives $4x = 4 \Rightarrow x = 1$, then $y = 1$

$\Rightarrow$ Point: $(1, 1)$

2. On y-axis $(x = 0)$:

• From $3x + 2y = 5 \Rightarrow y = 2.5$

• From $7x + 2y = 9 \Rightarrow y = 4.5$

$\Rightarrow$ Points: $(0, 2.5)$ and $(0, 4.5)$

So, vertices are:

$(0, 2.5), (0, 4.5), (1, 1)$

Now evaluate $Z = 4x + 6y$:

• At $(0, 2.5): Z = 15$
• At $(0, 4.5): Z = 27$ (maximum)
• At $(1, 1): Z = 10$ (minimum)

Thus:

• $(a, b) = (0, 4.5)$
• $(p, q) = (1, 1)$

Now compute:

$\frac{a}{p} + \frac{b}{q} = \frac{0}{1} + \frac{4.5}{1} = 4.5$

The correct answer is Option (2) $\rightarrow 4.5$