Question

Minimise Z = 3 x + 5 y such that .

Answer 1

The given constraints are:

x+3y≥3 x+y≥2 x≥0 y≥0

The given objective function which need to be minimize is,

Z=3x+5y

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

x	0	3
y	1	0

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

0+0≥3 0≥3

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

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

x	0	2
y	2	0

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

0+0≥2 0≥2

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

By the substitution method, the intersection points of the lines x+3y≥3 and x+y≥2 is ( 3 2 , 1 2 ).

Plot the points of all the constraint lines,

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

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

Corner points	Z=3x+5y
A( 3,0 )	9
B( 3 2 , 1 2 )	7 (minimum)
C( 0,2 )	10

As it is seen from the graph that the feasible region is unbounded, so, it can be said that 7 may or may not be the minimum value of Z.

Plot the graph of inequality 3x+5y<7 to find the feasibility.

From graph, it is clear that feasible region has no common points with the equation of inequality.

Therefore, the minimum value of Z is 7 at ( 3 2 ' 1 2 ).