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:
Answer & explanation
Correct answer: option 2
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) $\rightarrow 4.5$