For the objective function $Z=3x+5y$ subject to constraints $x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0$:

there exists only minimum value of Z

The correct answer is Option (2) → there exists only minimum value of Z

Objective function: $Z = 3x + 5y$

Constraints: $x + 3y \ge 3$, $x + y \ge 2$, $x \ge 0$, $y \ge 0$

The feasible region is unbounded in the first quadrant in the direction of increasing $Z$ because as $x$ and $y$ increase, $Z = 3x + 5y$ can become arbitrarily large.

However, there is a minimum value at the corner point nearest to origin satisfying constraints:

Intersection of $x+3y = 3$ and $x+y = 2$: Solve $x + 3y =3$, $x + y =2$ → Subtract: $2y =1 \Rightarrow y = \frac{1}{2}$, $x = 2 - \frac{1}{2} = \frac{3}{2}$

Minimum $Z_{min} = 3*(3/2) + 5*(1/2) = 9/2 + 5/2 = 14/2 = 7$

There exists only minimum value of $Z$