Answer :

It is given in the question that,

Z = - x + 2y

We have to subject on the following equation:

(x, y) = (5, 0), (0, 5)

(x, y) = (6, 0), (3, 0)

We can clearly see that the feasible region is unbounded and the corner points of the feasible region are (3, 2), (4, 1) and (6, 0)

As per the table maximum value of Z is 1 but we can see that the feasible region is unbounded. Thus, 1 may or may not be the minimum value of Z

We will plot the graph of inequality,

Here we will see is the resulting half plane has common points with the feasible region or not. As per the graph we can observe that the feasible region have some points in common with - x + 2y > 1.

Hence, there is no maximum value of Z

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

Corner points of Mathematics - Exemplar

Using the method Mathematics - Board Papers

Refer to ExerciseMathematics - Exemplar

A dietician wisheRD Sharma - Volume 2

A merchant plans Mathematics - Board Papers

A dietician wisheMathematics - Board Papers

Maximise and MiniMathematics - Exemplar