Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Linear Programming

Question:

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

Options:

Max: 11, Min: 0

Max: 9, Min: $\frac{12}{7}$

Max: 11, Min: $\frac{12}{5}$

Max: 13, Min: $\frac{12}{7}$

Correct Answer:

Max: 9, Min: $\frac{12}{7}$

Explanation:

The correct answer is Option (2) → Max: 9, Min: $\frac{12}{7}$

Here, corner points are given as follows:

$R\left(\frac{7}{2}, \frac{3}{4}\right), Q\left(\frac{3}{2}, \frac{15}{4}\right), P\left(\frac{3}{13}, \frac{24}{13}\right) \text{ and } S\left(\frac{18}{7}, \frac{2}{7}\right).$

Now, evaluating the value of $Z$ for the feasible region $RQPS$.

Corner points

Value of $Z=x+2y$

$R\left(\frac{7}{2}, \frac{3}{4}\right)$

$Z = \frac{7}{2} + 2\left(\frac{3}{4}\right) = 5$

$Q\left(\frac{3}{2}, \frac{15}{4}\right)$

$Z = \frac{3}{2} + 2\left(\frac{15}{4}\right) = 9$ $\leftarrow$ Maximum

$P\left(\frac{3}{13}, \frac{24}{13}\right)$

$Z = \frac{3}{13} + 2\left(\frac{24}{13}\right) = \frac{51}{13}$

$S\left(\frac{18}{7}, \frac{2}{7}\right)$

$Z = \frac{18}{7} + 2\left(\frac{2}{7}\right) = \frac{22}{7}$ $\leftarrow$ Minimum

Hence, the maximum value of $Z$ is 9 at $\left(\frac{3}{2}, \frac{15}{4}\right)$ and the minimum value of $Z$ is $\frac{22}{7}$ at $\left(\frac{18}{7}, \frac{2}{7}\right)$.