hat point in the feasible region maximizes the objective function
constraints:
x>=0
y>=0
-x+3>=y
y<=1/3 x+1
Objective function: C=5x-4y
1. Region limited by :
x>=0
y>=0
x + y <= 3
is the interior of rectangle triangle
of summits (0,0), (0,3)and (3,0)
if we add the constraint
y <= 1/3 x + 1
it's the part in the triangle below this line :
the summits are (0,0) , (0,1) , (3,0)
and the intersection point of
line L of equation : y = x/3 + 1 and the hypotenuse
of the triangle (equation x+y = 3)
let's solve this :
3 - x = x/3 + 1
4x/3 = 2
x = 3/2 and y = 3/2
now the Criteria : C = 5x - 4y
are lines parallel to line of equation
5x - 4y = 0
or
y = (5/4)x
so C is maximum at an edge of the domain :
points are
O ( 0 ,0)
A( 3 , 0)
B ( 0 ; 1)
D ( 3/2 ; 3/2)
criteria is C = 5x - 4y
C (A) = 5*3 - 4*0 = 15
C(B) = 5*0 - 4*1
C(D) = 5* (3/2) - 4*(3/2) = 3/2
so C is max at point A(3 ; 0)
