Solution:
According to the solution of exercise 14,
we have Maximize 
subject to constrains
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 
and 
It can be seen that the shaded region is the feasible region whose corner points are 
, 150), 
.
Evaluating the value of 
, we have
![\[\begin{tabular}{|l|l|} \hline Corner points & Value of $\mathrm{Z}=2000 \mathrm{x}+120 \mathrm{y}$ \\ \hline $0(0,0)$ & $\mathrm{Z}=200(0)+120(0)=0$ \\ \hline $\mathrm{A}(200,0)$ & $\mathrm{Z}=200(200)+120(0)=40000$ \\ \hline $\mathrm{B}(150,150)$ & $\mathrm{Z}=200(150)+120(150)=48000$ \\ \hline $\mathrm{C}(100,200)$ & $\mathrm{Z}=200(100)+120(200)=44000$ \\ \hline $\mathrm{D}(0,100)$ & $\mathrm{Z}=200(0)+120(100)=12000$ \\ \hline \end{tabular}\]](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 
is 48000 at 
which means 150 sweaters of each type.