Solution:
According to the solution of exercise 11, we have Maximize 
subject to the constraints
(i)
(ii)
(iii)
(iv)
Construct a constrain table for the above
![\[\begin{tabular}{l} Table for (i) \\ \begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 10 & 0 \\ \hline $\mathrm{y}$ & 0 & 20 \\ \hline \multicolumn{1}{l}{ Table for (ii) } \\ \hline $\mathrm{x}$ & 12 & 0 \\ \hline $\mathrm{y}$ & 0 & 6 \\ \hline \end{tabular} \\ \hline Table for (iii) \\ \hline \begin{tabular}{|l|l|l|} \hline $\mathrm{x}$ & 15 & 0 \\ \hline y & 0 & 5 \\ \hline \end{tabular} \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-992551fc9f0563c3abbd298829a13691_l3.png "Rendered by QuickLaTeX.com")
Next, solving (i) and (ii) we obtain,
Therefore, the corner point is 
.
Solving (ii) and (iii) we obtain,
and the corner point is 
Lastly, solving (i) and (iii) we obtain,
(not included in the feasible region)
Here, 
is the feasible region.
As a result, the corner points are 
and 
.
Now evaluate the value of 
![\[\begin{tabular}{|l|l|} \hline Corner points & Corresponding value of $\mathrm{Z}=50 \mathrm{x}+60 \mathrm{y}$ \\ \hline $\mathrm{O}(0,0)$ & $\mathrm{Z}=50(0)+60(0)=0$ \\ \hline $\mathrm{A}(10,0)$ & $\mathrm{Z}=50(10)+60(0)=500$ \\ \hline $\mathrm{B}(28 / 3,4 / 3)$ & $\mathrm{Z}=50(28 / 3)+60(4 / 3)=1400 / 3+240 / 3$ \\ & $=1640 / 3=546.6$ \\ \hline $\mathrm{C}(6,3)$ & $\mathrm{Z}=50(6)+60(3)=480$ \\ \hline $\mathrm{D}(0,5)$ & $\mathrm{Z}=50(0)+60(5)=300$ \end{tabular}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3864d5b0d57680a7711c784b34427828_l3.png "Rendered by QuickLaTeX.com")
So, the maximum profit is Rs 
which is not possible for number of items in fraction. Thus, the maximum profit for the manufacture is Rs 480 at 
i.e. Type 
and Type 
.