Explanation:
(D) : Given,
\(z=18 x+10 y\) subject to
\(4 x+y \geq 20,2 x+3 y \geq 30, x, y \geq 0\)
Draw the lines \(4 x+y=20,2 x+3 y=30\)
i.e., \(\frac{x}{5}+\frac{y}{20}=1, \frac{x}{15}+\frac{y}{10}=1\)

\(B\) is the point of intersection of the lines \(4 x+y=20\)
\(2 \mathrm{x}+3 \mathrm{y}=30\) i.e., \(\mathrm{B}=(3,8)\)
We have corner points \(\mathrm{A}(15,0), \mathrm{B}(3,8)\) and \(\mathrm{C}(0,20)\).
Now, \(z=18 x+10 y\)
Therefore, \(z(A)=18(15)+10(0)=270\)
\(z(B)=18(3)+10(8)=134\)
\(z(C)=18(0)+10(20)=200\)
Therefore, \(\mathrm{z}\) has minimum value 134