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!
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
To find the point in the feasible region that maximizes the objective function C = 5x - 4y, we need to first determine the feasible region and then evaluate the objective function at each point within the feasible region.
Feasible region:
Let's examine the given constraints one by one:
1. x >= 0: This constraint means that x must be greater than or equal to zero. It defines the boundary on the left side of the coordinate plane.
2. y >= 0: This constraint means that y must be greater than or equal to zero. It defines the boundary at the bottom of the coordinate plane.
3. -x + 3 >= y: This constraint represents the line with a slope of -1 passing through the point (3, 0). To find the points in the feasible region, we need to shade the region above this line.
4. y <= 1/3x + 1: This constraint represents a line with a slope of 1/3 passing through the point (-3, 1). To find the points in the feasible region, we need to shade the region below this line.
Graphically, the feasible region is the intersection of the shaded regions (above the line -x + 3 >= y and below the line y <= 1/3x + 1).
To find the points, we can solve the system of equations formed by the lines -x + 3 = y and y = 1/3x + 1.
Solving the system of equations:
Substitute y from equation 2 into equation 1 to eliminate y:
-x + 3 = 1/3x + 1
Multiply through by 3 to eliminate the fraction:
-3x + 9 = x + 3
Simplify and solve for x:
-4x = -6
x = 3/2
Substitute the value of x back into equation 2 to find y:
y = 1/3(3/2) + 1
y = 5/6
So, one of the points in the feasible region is (3/2, 5/6).
Next, we need to evaluate the objective function C = 5x - 4y at this point:
C = 5(3/2) - 4(5/6)
C = 15/2 - 20/6
C = 45/6 - 20/6
C = 25/6
Therefore, the value of the objective function at the point (3/2, 5/6) within the feasible region is 25/6.
Now, we need to evaluate the objective function at the extreme points of the feasible region. To do this, we can find the intersecting points of the lines that form the boundaries of the feasible region using the same method as above.
Once we have the extreme points, we can evaluate the objective function at each point and determine which one maximizes the value of C = 5x - 4y.