Given the system of constraints, name all vertices of the feasible region. Then find the maximum value of the given objective function.
constraints
x≥0
y≥0
y≤1/3x+3
5≥x+y
Objective Function:
C=6x−4y
x≥0 , y≥0 simply tells you that you are in quadrant I
consider y = 1/3 x + 3 , and 5 = x+y
sub the first into the 2nd:
5 = x + x/3 + 3
4/3 x = 2
x = 3/2 , then y = (1/3)(3/2) + 3 = 1/2 + 3 = 7/2
the intersection point is ((3/2 , 7/2)
for x+y = 5, x-intercept is (5,0), y-intercept is (0,5)
for y = x/3 + 3, the x-intercept is negative, thus of no interest to us, the y-intercept is (0,3)
normally your objective Function would have a negative slope, was expecting
C = 6x + 4y