The maximum value of z = 4x + 2y subject to constraints
2x + 3y ≤ 28,
x + y ≤ 10,
x, y ≥ 0 is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 40
Since x, y ≥ 0, the feasible region lies in the first quadrant, on the origin side of both lines (the origin satisfies both constraints).
The line 2x + 3y = 28 meets the axes at (14, 0) and \(\left(0, \frac{28}{3}\right)\); the line x + y = 10 meets them at (10, 0) and (0, 10). Both constraints must hold, so the region ends at (10, 0) on the x-axis and at \(\left(0, \frac{28}{3}\right)\) on the y-axis.
Where the two lines meet, put x = 10 − y in 2x + 3y = 28:
\(2(10 - y) + 3y = 28\)
\(20 + y = 28\)
\(y = 8,\quad x = 2\)
Values of z = 4x + 2y at the corner points:
\(z(0, 0) = 0\)
\(z(10, 0) = 40\)
\(z(2, 8) = 8 + 16 = 24\)
\( \displaystyle z\left(0, \frac{28}{3}\right) = \frac{56}{3} \)
The largest value is 40, at (10, 0). So the maximum value of z is 40.