Raymond B. answered 04/29/22
Math, microeconomics or criminal justice
x1+2x2 =10
6x1+6x2=36
x1 +x2 =6
subtract 3rd equation from 1st to get
x2 =4
x1 = 2
the above 2 boundary lines intersect at (2,4)
check that point as well as (0,5) and (4,0)
z = 4(2) + 5(4) = 8+20= 28
z = 0 + 5(5) = 25
z = 4(4) + 0 = 25
max z = 28 when x1=2 and x2 =4
graph the boundary lines of the inequalities, 2 downward sloping lines with x1 on the horizontal axis and x2 on the vertical axis. maximum point on the vertical axis is (0,5), maximum x1 coordinate on the horizontal axis is (4,0)
they intersect at (2,4) also graph 4x1+5x2 = 28 which intersects (2,4), (0,28/5) and (7,0)
solution has to be in quadrant I given the x1,x2>0 restriction
this problem resembles a comparable economics problme, the production possiblities frontier, although it uses a curve in place of the inequalities rather than two straight lines.
Ref V.
Therefore the feasible region is below the line 6x1+6x2 = 36? Thank you04/30/22