Question

Minimise and Maximise Z = 5 x + 10 y subject to .

Answer 1

The given constraints are,

x+2y≤120 x+y≥60 x−2y≥0 x≥0 y≥0

The given objective function which needs to maximise and minimise is,

Z=5x+10y

The line x+2y≤120 gives the intersection point as,

x	0	120
y	60	0

Also, when x=0,y=0 for the line x+2y≤120, then,

0+0≤120 0≤120

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

The line x+y≥60 gives the intersection point as,

x	0	60
y	60	0

Also, when x=0,y=0 for the line x+y≥60, then,

0+0≥60 0≥60

This is false, so the graph have the shaded region away the origin.

By the substitution method, the intersection points of the lines x+2y≤120 and x+y≥60 is ( 0,60 ).

The line x−2y≥0 gives the intersection point as,

x	0	0
y	0	0

Also, when x=0,y=0 for the line x−2y≥0, then,

0+0≥0 0≥0

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≤120 and x−2y≥0 is ( 60,30 ).

By the substitution method, the intersection points of the lines x−2y≥0 and x+y≥60 is ( 40,20 ).

Plot the points of all the constraint lines,

It can be observed that the corner points are A( 60,0 ),B( 120,0 ),C( 60,30 ),D( 40,20 )

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

Corner points	Z=5x+10y
A( 60,0 )	300 (Minimum)
B( 120,0 )	600 (Maximum)
C( 60,30 )	600 (Maximum)
D( 40,20 )	400

Therefore, the minimum value of Z is 300 at point ( 60,0 )and the maximum value of Z is 600 at all the points ( 120,0 ) and ( 60,30 ).