class 12 maths linear programming

The feasible region for a LPP is shown in the following figure. Evaluate $Z = 4x + y$ at each of the corner points of this region. Find the minimum value of Z, if it exists.

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📘 Linear Programming NCERT,Exemp,Q.9,Page,251 SA

The feasible region for a LPP is shown in the following figure. Evaluate $Z = 4x + y$ at each of the corner points of this region. Find the minimum value of Z, if it exists.

Official Solution

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From the shaded region, it is clear that feasible region is unbounded with the

corner points A(4,0), B(2,1) and C(0,3).

Also, We have the following conditions as per the question, $Z = 4x + y$.

[since, $x + 2y = 4$ and $x + y = 3 \Rightarrow y = 1$ and $x = 2$]

figure

Now, we see that 3 is the smallest value of Z at the corner point (0,3).

Note that here we see that, the region is unbounded, therefore 3 may or may not be the minimum value of Z.

To decide this issue, we graph the inequality $4x + y < 3$ and check whether the resulting open half plan has no point in common with feasible region otherwise,

Z has no minimum value.

From the shown graph above, it is clear that there is no point in common with feasible region and hence Z has minimum value 3 at (0,3).

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