Linear Inequalities

Concept:Find the corner points of the feasible region defined by the given linear inequalities.Explanation:The inequalities are $2x+5y \leq 20$, $3x+2y \leq 12$, $x \geq 0$, $y \geq 0$.The boundary lines are:$2x+5y=20$ with intercepts $(10,0)$ and $(0,4)$.$3x+2y=12$ with intercepts $(4,0)$ and $(0,6)$.The feasible region is in the first quadrant.The intersection of the two lines is found by solving:$2x+5y=20$ and $3x+2y=12$.Multiply first by 2: $4x+10y=40$; second by 5: $15x+10y=60$.Subtract: $11x=20 \implies x=\frac{20}{11}$. Then $2\cdot\frac{20}{11}+5y=20 \implies \frac{40}{11}+5y=20 \implies 5y=\frac{180}{11} \implies y=\frac{36}{11}$.So intersection is $\left(\frac{20}{11},\frac{36}{11}\right)$.Check which intercepts satisfy both inequalities:$(10,0)$ fails $3x+2y \leq 12$ ($30>12$), so not a vertex.$(0,6)$ fails $2x+5y \leq 20$ ($30>20$), so not a vertex.Valid vertices are $(0,0)$, $(0,4)$, $(4,0)$, and the intersection $\left(\frac{20}{11},\frac{36}{11}\right)$.Answer:Option A lists exactly these four points.