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:

2.5

The correct answer is Option (2) → 4.5

Constraints:

$3x + 2y \ge 5 \Rightarrow y \ge \frac{5-3x}{2}$

$7x + 2y \le 9 \Rightarrow y \le \frac{9-7x}{2}$

$x \ge 0, \ y \ge 0$

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

$3x+2y=5 \Rightarrow y = \frac{5-3x}{2}$

$7x+2y=9 \Rightarrow y = \frac{9-7x}{2}$

Equate: $\frac{5-3x}{2} = \frac{9-7x}{2} \Rightarrow 4x = 4 \Rightarrow x=1, y=1$

Check intersections with axes:

$3x+2y=5$: x=0 → y=5/2, y=0 → x=5/3

$7x+2y=9$: x=0 → y=9/2, y=0 → x=9/7

Feasible vertices satisfying all constraints: $(1,1)$ and $(0,5/2)$

Evaluate $Z = 4x + 6y$:

$(1,1) \Rightarrow Z = 4+6=10$ (minimum)

$(0,5/2) \Rightarrow Z = 0+15=15$ (maximum)

Maximum at $(a,b)=(0,5/2)$, Minimum at $(p,q)=(1,1)$

Compute $a/p + b/q = 0/1 + (5/2)/1 = 5/2$

Answer: $\frac{5}{2}$