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!
To find the point in the feasible region that maximizes the objective function, we need to find the corner points of the feasible region and evaluate the objective function at each of these points.
First, let's graph the constraints:
Constraint 1: x ≥ 0
This is a vertical line passing through the y-axis at x = 0.
Constraint 2: y ≥ 0
This is a horizontal line passing through the x-axis at y = 0.
Constraint 3: -x + 3 ≥ y
Rearranging this inequality, we have y ≤ -x + 3. So the line y = -x + 3 is our constraint.
Constraint 4: y ≤ 1/3x + 1
This is a line with a slope of 1/3 and a y-intercept of 1.
Now let's find the intersection points of these lines to determine the corner points:
At x = 0, y = -x + 3, we have y = 3.
So one corner point is (0, 3).
At y = 0, -x + 3 ≥ y gives us -x + 3 ≥ 0, which means x ≤ 3.
So another corner point is (3, 0).
At the intersection of y = -x + 3 and y = 1/3x + 1, we can solve for x:
-x + 3 = 1/3x + 1
Multiplying by 3 to clear the fraction, we get:
-3x + 9 = x + 3
Rearranging and simplifying:
4x = -6
x = -6/4
x = -3/2
Plugging this value back into y = -x + 3, we get:
y = -(-3/2) + 3
y = 3/2 + 3/2
y = 3
So another corner point is (-3/2, 3).
Now we evaluate the objective function at each corner point:
At (0, 3):
c = 5(0) - 4(3) = -12
At (3, 0):
c = 5(3) - 4(0) = 15
At (-3/2, 3):
c = 5(-3/2) - 4(3) = -7.5 - 12 = -19.5
Comparing these values, we can see that the maximum value of the objective function occurs at (3, 0), with a value of 15.