What 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
To find the point in the feasible region that maximizes the objective function, we need to first graph the feasible region and then find the corner point with the highest value of the objective function.
Let's start by graphing the constraints:
1. Plot the line x = 0 (vertical line through the origin).
2. Plot the line y = 0 (horizontal line through the origin).
3. Plot the line -x + 3 = y (diagonal line with y-intercept 3 and slope -1).
4. Plot the line y = (1/3)x + 1 (diagonal line with y-intercept 1 and slope 1/3).
Now let's find the intersection points of the lines:
- x + 3 = y (Equation of the line -x + 3 = y)
- x = y - 3 (Rearrange the equation)
- 0 = (1/3)x + 1 (Equation of the line y = (1/3)x + 1)
- (1/3)x = - 1 (Subtract 1 from both sides)
- x = 3 (Multiply both sides by 3)
- y = (1/3)(3) + 1 (Substitute x = 3 into equation y = (1/3)x + 1)
- y = 1 + 1 (Multiply 1/3 and 3)
- y = 2 (Add 1 and 1)
- x = 0 (Substitute y = 2 into equation x = y - 3)
- x = - 1 (Add -3 and 2)
The intersection points are (0, 2) and (-1, 3).
Now we check the objective function at each corner point:
C = 5x - 4y
For point (0, 2):
C = 5(0) - 4(2)
C = -8
For point (-1, 3):
C = 5(-1) - 4(3)
C = -17
Comparing the objective function values, we see that the point (0, 2) maximizes the objective function.
Therefore, the point (0, 2) in the feasible region maximizes the objective function.