The point which provides the optimal solution of the linear programming problem
maximize \(z =21x+35y\)
\(3x + 2y \le 30\)
\(4x+5y \le 60\)
\(x\ge 0,y\ge 0\)
has the coordinates
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \((0,12)\)
The corner points of the feasible region are found from the boundary lines. On the x-axis the first constraint is the tighter one (\(x \le 10\)); on the y-axis the second is tighter (\(y \le 12\)). The two lines meet where
\(3x + 2y = 30\)
\(4x + 5y = 60\)
Multiplying the first by 5 and the second by 2 and subtracting:
\(15x - 8x = 150 - 120\)
\(7x = 30\)
\( \displaystyle x = \frac{30}{7},\quad y = \frac{30 - \frac{90}{7}}{2} = \frac{60}{7} \)
The value of \(z = 21x + 35y\) at each corner point:
\((0, 0):\ z = 0\)
\((10, 0):\ z = 210\)
\( \displaystyle \left(\frac{30}{7}, \frac{60}{7}\right):\ z = 90 + 300 = 390 \)
\((0, 12):\ z = 420\)
The maximum, 420, is at \((0, 12)\).