For the objective function \(Z=4x+6y\) subject to the constraints \(3x + 2y \ge 5, 7x + 2y \le 9, x \ge 0, y \ge 0\), the maximum value of Z occurs at (a, b) and the minimum value of Z occurs at (p, q) then the value of \(\displaystyle\frac{a}{p}+\frac{b}{q}\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4.5
The optimum of a linear objective lies at a corner point of the feasible region. The two lines meet where
\((7x + 2y) - (3x + 2y) = 9 - 5\)
\(4x = 4, \quad x = 1, \quad y = 1\)
On the y-axis (x = 0), the constraints give \(y \ge 2.5\) and \(y \le 4.5\), so (0, 2.5) and (0, 4.5) are corners. On the x-axis (y = 0) they need \(x \ge \frac{5}{3}\) and \(x \le \frac{9}{7}\), which is impossible, so the region does not reach the x-axis.
Corner points and values of Z = 4x + 6y:
\((0, 2.5): Z = 15\)
\((0, 4.5): Z = 27 \text{ (maximum)}\)
\((1, 1): Z = 10 \text{ (minimum)}\)
So (a, b) = (0, 4.5) and (p, q) = (1, 1):
\( \displaystyle \frac{a}{p} + \frac{b}{q} = \frac{0}{1} + \frac{4.5}{1} = 4.5 \)