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(A) $1$. Consider the inequality $2x + y \leq 12$. The boundary line is $2x + y = 12$. For $x=0, y=12$; for $y=0, x=6$. The region includes the origin

Vedclass · Accepted Answer

The feasible region is a bounded polygon with vertices $(0,0), (6,0), (5,1), (0,3.5)$.

Vedclass · Answer

(A) $1$. Consider the inequality $2x + y \leq 12$. The boundary line is $2x + y = 12$. For $x=0, y=12$; for $y=0, x=6$. The region includes the origin $(0,0)$ since $0 \leq 12$.$2$. Consider the inequality $x + 2y \leq 7$. The boundary line is $x + 2y = 7$. For $x=0, y=3.5$; for $y=0, x=7$. The region includes the origin $(0,0)$ since $0 \leq 7$.$3$. The constraints $x \geq 0$ and $y \geq 0$ restrict the solution to the first quadrant.$4$. To find the intersection of $2x + y = 12$ and $x + 2y = 7$,multiply the second equation by $2$: $2x + 4y = 14$. Subtracting the first from this gives $3y = 2$,so $y = 2/3$. Then $x = 7 - 2(2/3) = 7 - 4/3 = 17/3$.$5$. The vertices of the feasible region are $(0,0), (6,0), (17/3, 2/3), (0, 3.5)$.

Solve the following system of inequalities graphically: $2x + y \leq 12, x + 2y \leq 7, x \geq 0, y \geq 0$.

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