For a feasible region bounded by the corner points $(0,3)$, $(3,2)$, and $(5,0)$ in the first quadrant, the non-trivial constraints of the linear programming problem are:
Show Hint
You can quickly check options by plugging in the intersection corner point $(3,2)$. For option (A): $3+2=5 \le 5$ (True) and $3+3(2)=9 \le 9$ (True).
Concept:
To find the inequalities corresponding to a given set of corner points forming a boundary line:
• Identify pairs of corner points that lie on the boundary lines.
• Find the equation of the lines passing through those points using the two-point form: $y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$.
• Use the origin $(0,0)$ test to identify the correct inequality sign since the region is toward the origin.
Step 1: Find the equation of the first line boundary.
Let us inspect the corner points. The point $(5,0)$ lies on the $x$-axis and $(3,2)$ is an adjacent corner. Let's find the line passing through $(3,2)$ and $(5,0)$:
\[
\text{Slope } m_1 = \frac{0 - 2}{5 - 3} = \frac{-2}{2} = -1
\]
Using point-slope form with $(5,0)$:
\[
y - 0 = -1(x - 5) \quad \Rightarrow \quad y = -x + 5 \quad \Rightarrow \quad x + y = 5
\]
Since the region contains the origin $(0,0)$, substituting $(0,0)$ gives $0 + 0 \le 5$, which means the inequality is:
\[
x + y \le 5
\]
Step 2: Find the equation of the second line boundary.
Now look at the remaining corner points: $(0,3)$ on the $y$-axis and $(3,2)$. Let us find the line passing through them:
\[
\text{Slope } m_2 = \frac{2 - 3}{3 - 0} = \frac{-1}{3}
\]
Using the $y$-intercept form ($c=3$):
\[
y = -\frac{1}{3}x + 3
\]
Multiplying the entire equation by 3:
\[
3y = -x + 9 \quad \Rightarrow \quad x + 3y = 9
\]
Testing the origin $(0,0)$ again, $0 + 3(0) \le 9$ is true, so the inequality constraint is:
\[
x + 3y \le 9
\]
Step 3: Combine constraints.
The non-trivial constraints forming this region in the first quadrant are:
\[
x + y \le 5 \quad \text{and} \quad x + 3y \le 9
\]
This matches option (A).
