If $\mathrm{Z}=10 x+25 y$ subject to $0 \leq x \leq 3,0 \leq y \leq 3, x+y \leq 5, x \geq 0, \mathrm{y} \geq 0$ then $\mathrm{z}$ is maximum at the point

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Aicharya · Accepted Answer

First we draw the lines $A B, C D$ and $E F$ whose equation are $x=3, y=3$ and $x+y=5$ respectively. \begin{array}{|c|c|c|c|c|c|} \hline Line & Equation & Points on the X-axis & Points on the Y-axis & Sign & Region \ \hline \mathrm{AB} & \mathrm{x}=3 & \mathrm{~A}(3,0) & - & \leq & origin side of line AB \ \hline \mathrm{CD} & \mathrm{y}=3 & - & \mathrm{D}(0,3) & \leq & origin side of line CD \ \hline \mathrm{EF} & \mathrm{x}+\mathrm{y}=5 & \mathrm{E}(5,0) & \mathrm{F}(0,5) & \leq & origin side of line EF \ \hline \end{array} The feasible region is OAPQDO which is shaded in the graph. The vertices of the feasible region are $\mathrm{O}(0,0), \mathrm{A}(3,0), \mathrm{P}, \mathrm{Q}$ and $\mathrm{D}(0,3)$. $P$ is the point of intersection of the lines $x+y=5$ and $x=3 \Rightarrow P \equiv(3,2)$ $Q$ is the point of intersection of the lines $x+y=5$ and $y=3 \Rightarrow Q=(2,3)$ The values of the objective function $Z=10 x+25 y$ at these vertices are $Z_{(0)}=10(0)+25(0)=0+0=0$ $Z_{(A)}=10(3)+25(0)=30+0=30$ $Z_{(P)}=10(3)+25(2)=30+50=80$ $Z_{(Q)}=10(2)+25(3)=20+75=95$ $Z_{(D)}=10(0)+25(3)=0+75=75$ $	herefore Z$ has maximum value 95, when $x=2$ and $y=3$.

If $\mathrm{Z}=10 x+25 y$ subject to $0 \leq x \leq 3,0 \leq y \leq 3, x+y \leq 5, x \geq 0, \mathrm{y} \geq…

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