(A)
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
$Z_{(0)}=10(0)+25(0)=0+0=0$
$Z_{(A)}=10(3)+25(0)=30+0=30$
$Z_{(P)}=10(3)+25(2)=30+50=80$
$Z_{(Q)}=10(2)+25(3)=20+75=95$
$Z_{(D)}=10(0)+25(3)=0+75=75$
$\therefore Z$ has maximum value 95, when $x=2$ and $y=3$.