The maximum value of \(Z = 15x + 10y\) subject to the constraints \(4x + 6y \le 360\), \(3x \le 180\), \(5y \le 200\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1100
The constraints are
\(2x + 3y \le 180\)
\(x \le 60\)
\(y \le 40\)
\(x \ge 0,\ y \ge 0\)
The corner points:
\(x = 60\) and \(2x + 3y = 180\) meet at (60, 20)
\(y = 40\) and \(2x + 3y = 180\) meet at (30, 40)
together with (0, 0), (60, 0) and (0, 40). Values of Z:
\(Z(0, 0) = 0\)
\(Z(60, 0) = 900\)
\(Z(60, 20) = 900 + 200 = 1100\)
\(Z(30, 40) = 450 + 400 = 850\)
\(Z(0, 40) = 400\)
The region is bounded, so the maximum value of Z is 1100, at (60, 20).