Consider an LPP: Maximise \(Z = 50x + 15y\) subjected to constraints \(x + y \le 60,\ 5x + y \le 100,\ x, y \ge 0\). If the maximum value of \(Z\) occurs at \(x = \alpha\) and \(y = \beta\), then the value of \(\alpha + \beta\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 60
The line \(x + y = 60\) meets the axes at (60, 0) and (0, 60), and \(5x + y = 100\) meets them at (20, 0) and (0, 100). Both constraints are of the form ≤, so the feasible region lies below both lines, in the first quadrant.
The two lines meet where
\((5x + y) - (x + y) = 100 - 60\)
\(4x = 40\)
\(x = 10,\ y = 50\)
So the corner points are (0, 0), (20, 0), (10, 50) and (0, 60). The value of Z at each:
\(Z(0, 0) = 0\)
\(Z(20, 0) = 1000\)
\(Z(10, 50) = 500 + 750 = 1250\)
\(Z(0, 60) = 900\)
The maximum is 1250 at (10, 50), so \(\alpha = 10\) and \(\beta = 50\):
\(\alpha + \beta = 60\)