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  • In Fig. 12.11, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y

In Fig. 12.11, the feasible region (shaded) for a LPP is shown. Determine the maximum and minimum value of Z = x + 2y

 

Answers (1)

It is given that:

Z = x + 2y

The figure that is given above, from that we can see that the shaded region is feasible and is bounded, so the maximum and the minimum point will occur at the corner point.

\begin{aligned} &\text { Corner Points are } \mathrm{P}\left(\frac{3}{13}, \frac{24}{13}\right), \mathrm{Q}\left(\frac{3}{2}, \frac{15}{4}\right), \mathrm{R}\left(\frac{7}{2}, \frac{3}{4}\right) \text { and } \mathrm{S}\left(\frac{18}{7}, \frac{2}{7}\right)\\ &\text { Now we will substitute these values in } \mathrm{Z} \text { at each of these corner points, we get } \end{aligned}

\begin{aligned} &\begin{array}{|l|l|} \hline \begin{array}{l} \text { Corner } \\ \text { Point } \end{array} & \text { Value of } \mathrm{Z}=\mathrm{x}+2 \mathrm{y} \\ \hline P\left(\frac{3}{13}, \frac{24}{13}\right) & 2=\left(\frac{3}{13}\right)+2\left(\frac{24}{13}\right)=\frac{3}{13}+\frac{48}{13}=\frac{51}{13} \\ Q\left(\frac{3}{2}, \frac{15}{4}\right) & 2=\left(\frac{3}{2}\right)+2\left(\frac{15}{4}\right)=\frac{3}{2}+\frac{15}{2}=\frac{18}{2}=9 \rightarrow \max \\ R\left(\frac{7}{2}, \frac{3}{4}\right) & 2=\left(\frac{7}{2}\right)+2\left(\frac{3}{4}\right)=\frac{7}{2}+\frac{3}{2}=\frac{10}{2}=5 \\ S\left(\frac{18}{7}, \frac{2}{7}\right) & 2=\left(\frac{18}{7}\right)+2\left(\frac{2}{7}\right)=\frac{18}{7}+\frac{4}{7}=\frac{22}{7}=3 \frac{1}{7} \rightarrow \min \\ \hline \end{array}\\ \end{aligned}

\\$ Hence, the maximum value of Z is 9 at the point $\left(\frac{3}{2}, \frac{15}{4}\right) .$\\ And the minimum value of $Z$ is $3 \frac{1}{7}$ at the point $\left(\frac{18}{7}, \frac{2}{7}\right)$.

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