What point in the feasible region maximizes the objective function ?
The constraints are:
x>=0
y>=0
-x+3>=y
y<=1/3x+1
Object function is:
C=5x-4y
To find the point in the feasible region that maximizes the objective function, we need to evaluate the objective function at each of the corner points of the feasible region.
The corner points of the feasible region can be found by solving the system of equations formed by the constraints.
First, rewriting the constraints in terms of x and y:
1) x >= 0
2) y >= 0
3) -x + 3 >= y
4) y <= (1/3)x + 1
Now, we can graph the feasible region by plotting the following points and lines:
1) (0, 0) - x and y intercepts of the feasible region
2) (3, 0) - intersection of line -x + 3 = y and x = 3
3) (6, 2) - intersection of line y = (1/3)x + 1 and x = 6
4) (0, 1) - intersection of line -x + 3 = y and y = 1
Next, we evaluate the objective function at each corner point:
1) C = 5(0) - 4(0) = 0
2) C = 5(3) - 4(0) = 15
3) C = 5(6) - 4(2) = 22
4) C = 5(0) - 4(1) = -4
From the calculations, we can see that the point (6, 2) maximizes the objective function. So, the point (6, 2) in the feasible region maximizes the objective function.
To find the point in the feasible region that maximizes the objective function C=5x-4y, we need to first plot the constraints and then determine the corner points or vertices of the feasible region.
1. Let's start by graphing the constraints:
- Graph the line x=0 (vertical line that passes through x=0) and shade the region to the right of the line for x>=0.
- Graph the line y=0 (horizontal line that passes through y=0) and shade the region above the line for y>=0.
- Graph the line -x+3=y (or y=-x+3) and shade the region above the line.
- Graph the line y=1/3x+1 and shade the region below the line.
2. Determine the intersection points or vertices of the shaded region to find the corner points of the feasible region.
3. Plug the coordinates of each corner point into the objective function C=5x-4y and calculate the corresponding objective function value.
4. Compare the objective function values at each corner point to identify the maximum value.
The corner point with the highest objective function value will be the point in the feasible region that maximizes the objective function.
To find the point in the feasible region that maximizes the objective function, we need to first determine the corner points of the feasible region.
The given constraints are:
1) x >= 0
2) y >= 0
3) -x + 3 >= y
4) y <= (1/3)x + 1
To find the corner points, we need to solve the equations formed by the intersection of these constraints.
First, let's consider the intersection of constraints 1) and 2). Since x and y are both non-negative, the corner point (0, 0) is obtained.
Next, let's find the intersection of constraints 1) and 3).
By setting -x + 3 = y, we can substitute this value for y in constraint 1) to get:
y <= (1/3)x + 1
Substituting -x + 3 for y:
-x + 3 <= (1/3)x + 1
Multiplying through by 3 to eliminate fractions:
-3x + 9 <= x + 3
Bringing x terms to one side and constant terms to the other:
-4x <= -6
Dividing by -4 (note that we reverse the inequality when dividing by a negative number):
x >= (3/2)
Since x has to be non-negative as per constraint 1), the corner point (3/2, 3/2) is obtained.
Now, let's find the intersection of constraints 1) and 4).
By setting y = (1/3)x + 1, we can substitute this value for y in constraint 1) to get:
y <= (1/3)x + 1
Substituting (1/3)x + 1 for y:
(1/3)x + 1 <= x
Subtracting (1/3)x from both sides:
1 <= (2/3)x
Multiplying through by 3/2 to eliminate fractions:
3/2 <= x
Since x has to be non-negative as per constraint 1), the corner point (3/2, 3/2) is obtained again.
Now, let's find the intersection of constraints 2) and 3).
By substituting y = -x + 3 in constraint 2), we get:
-x + 3 >= y
Substituting -x + 3 for y:
-x + 3 >= -x + 3
This means that this constraint is already satisfied by all values of x and y that satisfy the other constraints.
Finally, let's find the intersection of constraints 2) and 4).
By substituting y = (1/3)x + 1 in constraint 2), we get:
(1/3)x + 1 <= y
Substituting (1/3)x + 1 for y:
(1/3)x + 1 <= (1/3)x + 1
This means that this constraint is already satisfied by all values of x and y that satisfy the other constraints.
Now, we have the two corner points: (0, 0) and (3/2, 3/2).
To determine which of these corner points maximizes the objective function C = 5x - 4y, we can substitute the coordinates of each point into the objective function and compare the resulting values.
For (0, 0):
C = 5(0) - 4(0) = 0
For (3/2, 3/2):
C = 5(3/2) - 4(3/2) = 15/2 - 12/2 = 3/2
Therefore, the point (3/2, 3/2) in the feasible region maximizes the objective function.