The maximum value of the function z = 3x + 3y, subject to the constraints x + 2y ≤ 30, 2x + y ≤ 50, x ≥ 0, y ≥ 0 is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 80
The maximum of z is reached at a corner point of the feasible region.
The line x + 2y = 30 meets the axes at (30, 0) and (0, 15); the line 2x + y = 50 meets them at (25, 0) and (0, 50). With x, y ≥ 0, the feasible region has corners O(0, 0), A(25, 0), B(0, 15) and the point C where the two lines meet.
For C, from 2x + y = 50, y = 50 − 2x. Putting this in x + 2y = 30:
\(x + 100 - 4x = 30\)
\(-3x = -70\)
\( \displaystyle x = \frac{70}{3},\quad y = 50 - \frac{140}{3} = \frac{10}{3} \)
Values of z = 3x + 3y at the corners:
\(z(0, 0) = 0\)
\(z(25, 0) = 75\)
\(z(0, 15) = 45\)
\( \displaystyle z\left(\frac{70}{3}, \frac{10}{3}\right) = 3\left(\frac{70}{3} + \frac{10}{3}\right) = 80 \)
The maximum value is 80, at \(\left(\frac{70}{3}, \frac{10}{3}\right)\).
Why the other options are incorrect
Option (1) 75 — Incorrect. This is the value at the corner (25, 0), which is less than 80.
Option (2) 90 — Incorrect. This is z at (30, 0), which is not feasible: there 2x + y = 60, more than 50.
Option (4) 45 — Incorrect. This is the value at the corner (0, 15), which is less than 80.