Question

Minimise and Maximise Z = x + 2 y subject to .

Answer 1

The given constraints are,

x+2y≥100 2x−y≤0 2x+y≤200 x≥0 y≥0

The given objective function which needs to maximize and minimize is,

Z=x+2y

The line x+2y≥100 gives the intersection point as,

x	0	100
y	50	0

Also, when x=0,y=0 for the line x+2y≥100, then,

0+0≥100 0≥100

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

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

x	0	0
y	0	0

Also, when x=0,y=0 for the line 2x−y≤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≥100 and 2x−y≤0 is ( 20,40 ).

The line 2x+y≤200 gives the intersection point as,

x	0	100
y	200	0

Also, when x=0,y=0 for the line 2x+y≤200, then,

0+0≤200 0≤200

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≥100 and 2x+y≤200 is ( 100,0 ).

By the substitution method, the intersection points of the lines 2x+y≤200 and 2x−y≤0 is ( 50,100 ).

Plot the points of all the constraint lines,

It can be observed that the corner points are A( 0,50 ),B( 20,40 ),C( 50,100 ),D( 0,200 ).

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

Corner points	Z=x+2y
A( 0,50 )	100 (Minimum)
B( 20,40 )	100 (Minimum)
C( 50,100 )	250
D( 0,200 )	400 (Maximum)

Therefore, the maximum value of Z is 400 at ( 0,200 )and the minimum value of Z is 100 at the points ( 0,50 ) and ( 20,40 ).