x>=0, y>=0 simply says that we are staying in quadrant I
find the intersection of
y = -x + 3 and y = (1/3)x + 1
(1/3)x + 1 = -x + 3
x + 3 = -3x + 9
4x = 6
x = 3/2 or 1.5
then y = -1.5 + 3 = 1.5
sub into C = 5x - 4y to get
C = 5(1.5) - 4(1.5) = 1.5
You should also check if the x and y intercepts of each of the lines
give you a higher value of C
What point in the feasible region maximizes the objective function? Constraints:
x>=0
y>=0
-x+3>=y
y<=1/3x+1
Objective function: C=5x-4y
if u can just tell me what the points are and how u got them so i can understand better that would be great!
1 answer