4x + 3y = 24
Find the x- interecpt
substitute x = 0
4(0) + 3y = 24
3y = 24
y = 8
substitute y = 0
4x + 3(0) = 24
4x = 24
x =6
The intercepts are;
(0,8) and (6, 0)
x + 3y = 15
substitute x = 0
3y = 15
y =5
substitute y=0
x= 15
The intercept is;
(0, 5) and (15, 0)
Find the point of intersection by solving;
4x + 3y = 24 and x + 3y = 15 simultaneously
4x + 3y = 24 --------------------(1)
x + 3y = 15 --------------------(2)
subtract equation (2) from equation(1)
3x = 9
Divide bothside of the equation by 3
x = 3
substitute x = 3 into into equation (2)
3 + 3y = 15
subtract 3 from bothside
3y = 15 - 3
3y = 12
Divide both-side of the equation by 3
y = 4
(3, 4)
The coordinate of the feasible regions are;
(0, 8) (3, 4) and (15, 0)
Evaluate vertex values into the function below and solve for C
C= 6x + 7y
C = 6(0) + 7(8) = 0 + 56 = 56
C = 6(3) + 7(4) = 18 + 28 = 46
C = 6(15) + 7(0) = 90 + 0 = 90
Therefore the minimum value of C = 46 at (3, 4)