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

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

Constraints:

• $x + 3y \geq 3$
• $x + y \geq 2$
• $x \geq 0$
• $y \geq 0$

The constraints use $\geq$ signs, indicating the feasible region is above/right of the boundary lines.

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Finding the corner points where constraint lines intersect:

Intersection of $x + 3y = 3$ and $x + y = 2$:

From $x + y = 2$:

$x = 2 - y$

Substituting into $x + 3y = 3$:

$(2 - y) + 3y = 3$

$2 + 2y = 3$

$2y = 1$

$y = 0.5$

$x = 2 - 0.5 = 1.5$

Point A: $(1.5, 0.5)$

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Intersection of $x + 3y = 3$ and $y = 0$:

$x + 3(0) = 3$

$x = 3$

Point B: $(3, 0)$

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Intersection of $x + y = 2$ and $x = 0$:

$0 + y = 2$

$y = 2$

Point C: $(0, 2)$

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Since all constraints use $\geq$ and the feasible region extends infinitely upward and to the right, the region is unbounded.

The objective function $Z = 3x + 5y$ has positive coefficients. As $x$ or $y$ increase, $Z$ increases.

In an unbounded region where both variables can increase indefinitely, $Z$ has no maximum value.

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Evaluating $Z = 3x + 5y$ at each corner point:

At point A $(1.5, 0.5)$:

$Z = 3(1.5) + 5(0.5)$

$Z = 4.5 + 2.5$

$Z = 7$

At point B $(3, 0)$:

$Z = 3(3) + 5(0)$

$Z = 9$

At point C $(0, 2)$:

$Z = 3(0) + 5(2)$

$Z = 10$

The minimum value is $Z = 7$ at point $(1.5, 0.5)$.

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Therefore, there exists only a minimum value of $Z$.