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

(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.


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 \(y=3\).