Solution:
According to the solution of exercise 14,
we have Maximize 
Let’s construct a constrain table for the above
Table for (i)
![\[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 0 & 200 \\ \hline $\mathrm{y}$ & 600 & 0 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-54352ab21ba1244efd4b8ffbbb612b83_l3.png "Rendered by QuickLaTeX.com")
Table for (ii)
![\[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 0 & 300 \\ \hline $\mathrm{y}$ & 300 & 0 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cba15e09dc53ce4b08977597f3edae78_l3.png "Rendered by QuickLaTeX.com")
Table for (iii)
![\[\begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & $-100$ & 0 \\ \hline $\mathrm{y}$ & 0 & 100 \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-629e0bc372add10c71d093fe178c148a_l3.png "Rendered by QuickLaTeX.com")
On solving (i) and (ii), we obtain 
It can be seen that the shaded region is the feasible region whose corner points are 
Evaluating the value of 
![\[\begin{tabular}{|l|l|} \hline Corner points & Value of $\mathrm{Z}=2000 \mathrm{x}+120 \mathrm{y}$ \\ \hline $0(0,0)$ & $\mathrm{Z}=200(0)+120(0)=0$ \\ \hline $\mathrm{A}(200,0)$ & $\mathrm{Z}=200(200)+120(0)=40000$ \\ \hline $\mathrm{B}(150,150)$ & $\mathrm{Z}=200(150)+120(150)=48000$ \\ \hline $\mathrm{C}(100,200)$ & $\mathrm{Z}=200(100)+120(200)=44000$ \\ \hline $\mathrm{D}(0,100)$ & $\mathrm{Z}=200(0)+120(100)=12000$ \\ \hline \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-de047377f004cecf2aee08f1ec062e2a_l3.png "Rendered by QuickLaTeX.com")
It is seen from the above table that the maximum value is 48000 .
So, the maximum value of 
