Explanation:
(B) : Given, \(\mathrm{Z}=8 \mathrm{x}+3 \mathrm{y}\) subject to
\(\mathrm{x}+\mathrm{y} \leq 3\)
\(4 x+y \leq 6\)
\(\mathrm{x} \geq 0, \mathrm{y} \geq 0\)
using graphical method we draw the lines as follows :

So, the corner points are
\(\mathrm{A} \rightarrow(0,3)\)
\(\mathrm{B} \rightarrow\) In order to get point \(\mathrm{B}\), we solve
\(\mathrm{x}+\mathrm{y}=3\)
\(4 x+y=6\)
Solving the equation (i) and (ii),
\(3 x=3\)
\(\mathrm{x}=1, \quad \mathrm{y}=2\)
\(\therefore \mathrm{B}(1,2)\)
Point \(\mathrm{C}\) is \(\left(\frac{3}{2}, 0\right), \quad\) point \(\mathrm{D}\) is \((0,0)\)
Now, calculate the value of,
\(Z=8 x+3 y\)
So, at point \(\mathrm{A}(0,3) \mathrm{z}=9\)
Point B \((1,2) Z=14\)
Point \(\mathrm{C}\)
\(\left(\frac{3}{2}, 0\right) \mathrm{Z}=12\)
Point \(D(0,0) Z=0\)
So the max value is 14 which corresponds to \((1,2)\)