The point at which the maximum value of x + y, subject to constraints \(x+2y \le 70\), \(2x+y \le 95\), \(x \ge 0\), \(y \ge 0\) is obtained, is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (40, 15)
The maximum of z = x + y over the bounded feasible region is reached at a corner point.
The line x + 2y = 70 meets the axes at (70, 0) and (0, 35); the line 2x + y = 95 meets them at (47.5, 0) and (0, 95). So the corners on the axes are (0, 0), (47.5, 0) and (0, 35). The two lines meet where
\(x = 70 - 2y\)
\(2(70 - 2y) + y = 95\)
\(140 - 3y = 95,\quad y = 15,\quad x = 40\)
Values of z at the corners:
\(z(0, 0) = 0\)
\(z(47.5, 0) = 47.5\)
\(z(40, 15) = 55\)
\(z(0, 35) = 35\)
So the maximum value 55 is obtained at (40, 15).
Shortcut: The answer must be a feasible point. Checking x + 2y ≤ 70 for the options: (30, 25) gives 80, (20, 35) gives 90 and (35, 20) gives 75, all more than 70. Only (40, 15), with 40 + 30 = 70, satisfies it.