A firm makes x units of product P and y units of product Q a day, subject to \(x + 2y \le 20\) (labour hours), \(3x + 2y \le 36\) (raw material in kg), \(x \ge 0\), \(y \ge 0\). The profit is Rs. 30 on each unit of P and Rs. 20 on each unit of Q. Which of the following is true?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The maximum profit is Rs. 360, earned at every point of the segment joining (12, 0) and (8, 6).
Maximise \(Z = 30x + 20y\). The two constraint lines meet where, subtracting \(x + 2y = 20\) from \(3x + 2y = 36\),
\(2x = 16\)
\(x = 8,\ y = 6\)
The region is bounded, with corner points (0, 0), (12, 0), (8, 6) and (0, 10). Values of Z:
\(Z(0, 0) = 0\)
\(Z(12, 0) = 360\)
\(Z(8, 6) = 240 + 120 = 360\)
\(Z(0, 10) = 200\)
The maximum 360 occurs at two adjacent corners. Here
\(Z = 30x + 20y = 10(3x + 2y)\)
so the profit line is parallel to the raw-material line \(3x + 2y = 36\), and on that edge \(Z = 10 \times 36 = 360\). Every point of the segment joining (12, 0) and (8, 6) gives the maximum: there are infinitely many optimal solutions.