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(A) The corner points of the feasible region are $A, B, C, D$. From the graph,the lines are $y = 3$,$x = 4$,$y = x + 3$,and $2x + 3y = 12$. $1$. Point

Vedclass · Accepted Answer

$19.5$

Vedclass · Answer

(A) The corner points of the feasible region are $A, B, C, D$. From the graph,the lines are $y = 3$,$x = 4$,$y = x + 3$,and $2x + 3y = 12$. $1$. Point $A$ is the intersection of $y = 3$ and $2x + 3y = 12$: $2x + 3(3) = 12 \implies 2x = 3 \implies x = 1.5$. So,$A = (1.5, 3)$. $2$. Point $B$ is the intersection of $y = 3$ and $x = 4$. So,$B = (4, 3)$. $3$. Point $C$ is the intersection of $x = 4$ and $y = x + 3$: $y = 4 + 3 = 7$. So,$C = (4, 7)$. $4$. Point $D$ is the intersection of $y = x + 3$ and $2x + 3y = 12$: $2x + 3(x + 3) = 12 \implies 5x + 9 = 12 \implies 5x = 3 \implies x = 0.6$. $y = 0.6 + 3 = 3.6$. So,$D = (0.6, 3.6)$. Now,evaluate $Z = 3x + 5y$ at these points: $Z(A) = 3(1.5) + 5(3) = 4.5 + 15 = 19.5$. $Z(B) = 3(4) + 5(3) = 12 + 15 = 27$. $Z(C) = 3(4) + 5(7) = 12 + 35 = 47$. $Z(D) = 3(0.6) + 5(3.6) = 1.8 + 18 = 19.8$. The minimum value of $Z$ is $19.5$.

The shaded area in the figure given below is a solution set of a system of inequations. The minimum value of the objective function $Z = 3x + 5y$,subject to the linear constraints given by this system of inequations,is:

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