Question

Maximise Z = − x + 2 y , subject to the constraints: .

Answer 1

The given constraints are,

x≥3 x+y≥5 x+2y≥6 y≥0

The given objective function which needs to maximize is,

Z=−x+2y

The line x+y≥5 gives the intersection point as,

x	0	5
y	5	0

Also, when x=0,y=0 for the line x+y≥5, then,

0+0≥5 0≥5

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

The line x+2y≥6 gives the intersection point as,

x	0	6
y	3	0

Also, when x=0,y=0 for the line x+2y≥6, then,

0+0≥6 0≥6

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

By the substitution method, the intersection points of the lines x+y≥5 and x+2y≥6 is ( 4,1 ).

Plot the points of all the constraint lines,

It can be observed that the corner points are A( 6,0 ),B( 4,1 ),C( 3,2 ).

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

Corner points	Z=−x+2y
A( 6,0 )	-6
B( 4,1 )	-2
C( 3,2 )	1 (Maximum)

Since the feasible region is unbounded, so, 1 may or may not be the maximum value of Z.

Plot the graph of inequality −x+2y>7 to find the feasibility.

It can be observed that feasible region has common points with the equation of inequality, so Z=1 is not the maximum value.

Therefore, Z has no maximum value.