The maximum value of \(Z = -2x - 3y\) subject to the constraints \(\displaystyle\frac{x}{2} + \frac{y}{3} \le 1\), \(\displaystyle\frac{x}{3} + \frac{y}{2} \le 1\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(0\)
Multiplying each constraint by 6:
\(3x + 2y \le 6\)
\(2x + 3y \le 6\)
The region is bounded. Adding and subtracting the two equations of the boundary lines:
\(5x + 5y = 12\)
\(x - y = 0\)
\( \displaystyle x = y = \frac{6}{5} \)
The corner points are (0, 0), (2, 0), \(\left(\frac{6}{5}, \frac{6}{5}\right)\) and (0, 2). Values of Z:
\(Z(0, 0) = 0\)
\(Z(2, 0) = -4\)
\( \displaystyle Z\left(\frac{6}{5}, \frac{6}{5}\right) = -\frac{12}{5} - \frac{18}{5} = -6 \)
\(Z(0, 2) = -6\)
The largest value is 0, at the origin. This is expected: both coefficients of Z are negative, so Z is largest when x and y are as small as possible.