Question

The constraints of a problem are listed below. What are the vertices of the feasible region?

Answer 1

The vertices are the intersections between the lines.

1) line x + y = 5 and y = 3:

y = 3 => x + 3 = 5 => x = 5 - 3 => x = 2

=> vertix = (2,3)

2) line x + y = 5 and y = 0

=> y = 0 => x + 0 = 5 => x = 5

=> vertix = (5,0)

3) line x + y = 5 and x = 0

=> 0 + y = 5 => y = 5

=>  (0,5) ------> this is not a vertix because it is above the line y = 3

4) line y = 3 and x-axis

=> vertix = (0,3)

5) x-axis and y-axis => origin => vertix = (0,0)

So, there are four vertices: (0,0), (5,0), (2,3) and (0,3)

Answer: That is the first option:(0, 0), (0, 3), (2, 3), (5, 0)

Answer 2

Answer:

Option A is correct .i.e., ( 0, 0 ) , ( 0, 3) , ( 2, 3 ) & ( 5, 0)

Step-by-step explanation:

Given constraints are

x + y ≤ 5  ,  y ≤ 3  ,  x ≥ 0  &  y ≥ 0

To find Vertex of Feasible region we find point of intersection of lines.

To find Point of intersection we replace sign of inequality with equality.

So,

Point of intersection of  x + y = 5  ,  y = 3

put y = 3 in another equation, we get

x + 3 = 5

⇒ x = 5 - 3

⇒ x = 2

⇒ Point of intersection or vertex = ( 2 , 3 )

 

Point of intersection of  y = 3  ,  x = 0

⇒ Point of intersection or vertex = ( 0 , 3 )

Point of intersection of  x + y = 5  ,  y = 0

put y = 0 in another equation, we get

x + 0 = 5

⇒ x = 5  

⇒ Point of intersection or vertex = ( 5 , 0 )

Point of intersection of  x = 0  ,  y = 0

⇒ Point of intersection or vertex = ( 0 , 0 )

Point of intersection of  x + y = 5  ,  x = 0

put x = 0 in another equation, we get

0 + y = 5

⇒ y = 5  

⇒ Point of intersection = ( 0 , 5 ) but this is not required vertex as it lie on the above of y = 3

Therefore, Option A is correct .i.e., ( 0, 0 ) , ( 0, 3) , ( 2, 3 ) & ( 5, 0)