Maximize z=6x+4y subject to constraints,
`xle2,x+yle3,-2x+yle1,xyge0`
Also find the maximum value of 'z'.

Explanation

Consider x+y=3 " " -2x+y=1

Shaded region OPQRC is the feasible region with vertices `O(0,0),P(2,0),Q,RandC(0,1).Q` is a point of intersection of lines x=2 and x+y=3. Solving these equations, we get Q=(2,1) . R is the point of intersection of lines x+y=3 and -2x+y=1.
`x+y=3" "......(i)`
`-2x+y=1" "......(ii)`
`(+""--)/(3x=2)`
`:.x=(2)/(3)`
Substituting `x=(2)/(3)` in equation (i)
`(2)/(3)+y=3`
`:.y=3-(2)/(3)=(9-2)/(3)=(7)/(3)`
`:.R=((2)/(3),(7)/(3))`.
Now `z=6x+4y`
`:.z(0,0)=6xx0+4xx0=0`
`z(2,0)=6xx2+0=12`
`z(2,1)=6xx2+4xx1=16`
`z((2)/(3),(7)/(3))=6xx(2)/(3)+4xx(7)/(3)`
`=4+(28)/(3)=(40)/(3)=13.33`
`z(0,1)=6xx0+4xx1=4`
`:.` z has maximum value at (2,1)
`:.` Maximum value of z is 16 at x=2 and y=1.