For the objective function \(Z = 3x + 5y\) subject to constraints \(x + 3y \ge 3, x + y \ge 2, x \ge 0, y \ge 0\):
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → there exists only minimum value of Z
Both constraints are of the “≥” type, so the feasible region lies on the far side of the lines \(x + 3y = 3\) and \(x + y = 2\) from the origin and is unbounded. Its corner points are
\((0, 2)\): \(x + y = 2\) meets \(x = 0\) (there \(x + 3y = 6 \ge 3\))
\( \displaystyle \left(\frac{3}{2}, \frac{1}{2}\right) \): \(x + 3y = 3\) meets \(x + y = 2\) (subtracting, \(2y = 1\))
\((3, 0)\): \(x + 3y = 3\) meets \(y = 0\) (there \(x + y = 3 \ge 2\))
Values of Z at the corners:
\(Z(0, 2) = 10\)
\( \displaystyle Z\left(\frac{3}{2}, \frac{1}{2}\right) = \frac{9}{2} + \frac{5}{2} = 7 \)
\(Z(3, 0) = 9\)
No maximum: every point (x, 0) with \(x \ge 3\) is feasible, and there Z = 3x grows without limit.
Minimum exists: the smallest corner value is 7. Because the region is unbounded, check that Z never goes below 7. Writing Z through the two constraints:
\(3x + 5y = (x + 3y) + 2(x + y) \ge 3 + 2(2) = 7\)
So the open half-plane \(3x + 5y \lt 7\) has no point in common with the region, and the minimum value 7 occurs at \(\left(\frac{3}{2}, \frac{1}{2}\right)\). Hence Z has only a minimum value.