Question

Maximise Z = 3 x + 2 y subject to .

Answer 1

The given constraints are,

x+2y≤10 3x+y≤15 x≥0 y≥0

The objective function which needs to maximize is,

Z=3x+2y

The line x+2y≤10 gives the intersection point as,

x	0	1
y	5	0

Also, when x=0,y=0 for the line x+2y≤10, then,

0+0≤10 0≤10

This is true, so the graph have the shaded region towards the origin.

The line 3x+y≤15 gives the intersection point as,

x	0	5
y	15	0

Also, when x=0,y=0 for the line 3x+y≤15, then,

0+0≤15 0≤15

This is true, so the graph have the shaded region towards the origin.

By the substitution method, the intersection points of the lines x+2y≤10 and 3x+y≤15 is ( 4,3 ).

Plot the points of all the constraint lines,

It can be seen that the corner points are A( 5,0 ),B( 4,3 ),C( 0,5 ).

Substitute these points in the given objective function to find the minimum value of Z.

Corner points	Z=3x+2y
A( 5,0 )	15
B( 4,3 )	18 (maximum)
C( 0,5 )	10

Therefore, the maximum value of Z is 18 at the point ( 4,3 ).