The minimum value of \(Z = 7x + 5y\) subject to the constraints \(3x + 4y \le 24\), \(x \le 6\), \(x \ge 0\), \(y \ge 0\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0
The region lies below the line 3x + 4y = 24 and to the left of x = 6, in the first quadrant. It is bounded.
The line 3x + 4y = 24 meets the y-axis at (0, 6), and meets x = 6 where
\(4y = 24 - 18 = 6\)
\(\displaystyle y = \frac{3}{2}\)
The corner points are (0, 0), (6, 0), \(\left(6, \frac{3}{2}\right)\) and (0, 6).
\(Z(0, 0) = 0\)
\(Z(6, 0) = 42\)
\(\displaystyle Z\left(6, \frac{3}{2}\right) = 42 + \frac{15}{2} = \frac{99}{2}\)
\(Z(0, 6) = 30\)
The minimum is 0, at the origin. This is expected: both coefficients of Z are positive, so Z is never negative when \(x, y \ge 0\), and the origin is feasible.
Why the other options are incorrect
Option (1) 30 — Incorrect. This is Z at the corner (0, 6); Z at (0, 0) is smaller.
Option (3) 42 — Incorrect. This is Z at the corner (6, 0), which is not the smallest corner value.
Option (4) Z has no minimum value — Incorrect. The region is bounded, so Z has a minimum, namely 0.