Maximise Z = 5 x + 3 y subject to .
The given constraints are,
3 x + 5 y ≤ 15 5 x + 2 y ≤ 10 x ≥ 0 y ≥ 0
The given objective function which needs to maximize is,
Z = 5 x + 3 y
The line 3 x + 5 y ≤ 15 gives the intersection point as,
x 0 5 y 3 0
Also, when x = 0 , y = 0 for the line 3 x + 5 y ≤ 15 , then,
0 + 0 ≤ 15 0 ≤ 15
This is true, so the graph have the shaded region towards the origin.
The line 5 x + 2 y ≤ 10 gives the intersection point as,
x 0 2 y 5 0
Also, when x = 0 , y = 0 for the line 5 x + 2 y ≤ 10 , then,
0 + 0 ≤ 10 0 ≤ 10
This is true, so the graph have the shaded region towards the origin.
By the substitution method, the intersection points of the lines 3 x + 5 y ≤ 15 and 5 x + 2 y ≤ 10 is ( 20 19 , 45 19 ) .
Plot the points of all the constraint lines,
It can be observed that the corner points are O ( 0 , 0 ) , A ( 2 , 0 ) , B ( 0 , 3 ) , C ( 20 19 , 45 19 ) .
Substitute these points in the given objective function to find the maximum value of Z.
Corner points Z = 5 x + 3 y O ( 0 , 0 ) 0 A ( 2 , 0 ) 10 B ( 0 , 3 ) 9 C ( 20 19 , 45 19 ) 235 19 (Maximum)
Therefore, the maximum value of Z is 235 19 at the point ( 20 19 , 45 19 ) .
The given constraints are,
The given objective function which needs to maximize is,
The line
| x | 0 | 5 |
| y | 3 | 0 |
Also, when
This is true, so the graph have the shaded region towards the origin.
The line
| x | 0 | 2 |
| y | 5 | 0 |
Also, when
This is true, so the graph have the shaded region towards the origin.
By the substitution method, the intersection points of the lines
Plot the points of all the constraint lines,
It can be observed that the corner points are
Substitute these points in the given objective function to find the maximum value of Z.
| Corner points | |
| | 0 |
| | 10 |
| | 9 |
| | |
Therefore, the maximum value of Z is
.