The maximum value of \(P = 6x + 8y\) subject to the constraints \(2x + y \le 30\), \(x + 2y \le 24\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(120\)
The line \(2x + y = 30\) meets the x-axis at (15, 0), and \(x + 2y = 24\) meets the y-axis at (0, 12). The two lines meet where
\(4x + 2y = 60\)
\(x + 2y = 24\)
\(3x = 36\)
\(x = 12, \; y = 6\)
The bounded region has corner points (0, 0), (15, 0), (12, 6) and (0, 12). Values of P:
\(P(0, 0) = 0\)
\(P(15, 0) = 90\)
\(P(12, 6) = 72 + 48 = 120\)
\(P(0, 12) = 96\)
The maximum value is 120, at (12, 6).