Use the following constraints and objective function to answer the next 3 questions.
x >/ 0
y >/ 0
-x + 3 >/ y
y </ 1/3x + 1
Objective function: C=5x−4y
Question 1: Using a graphing calculator, find and select all the vertices for the feasible region.
Responses
(0,3)
(3,0)
(0,0)
(1.5,1.5)
(-3,0)
(1,0)
(0,1)
Question 2 Which vertex of the feasible region from above would maximize the objective function? (Make sure to show all necessary steps to support your answer.)
Question 3 What is the maximum value for C?
Question 1: Using a graphing calculator, find and select all the vertices for the feasible region.
To find the vertices of the feasible region, we need to graph the inequalities and determine where they intersect. Using a graphing calculator or software, we can plot the two inequality lines and shade the area that satisfies both conditions.
The graph shows that the feasible region is a triangle formed by three lines:
1. y ≥ -x + 3 (represented by the line in the first quadrant with slope -1 and y-intercept 3).
2. y ≤ 1/3x + 1 (represented by the line with slope 1/3 and y-intercept 1).
3. x ≥ 0 and y ≥ 0 (representing the positive x and y axes).
The vertices of the feasible region are the intersections of these lines.
From the given options, the vertices of the feasible region are:
(0, 3), (3, 0), and (1.5, 1.5)
Question 2: Which vertex of the feasible region from above would maximize the objective function? (Make sure to show all necessary steps to support your answer.)
To determine which vertex of the feasible region maximizes the objective function C = 5x - 4y, we need to evaluate the objective function at each vertex and compare the values.
1. C(0, 3) = 5(0) - 4(3) = -12
2. C(3, 0) = 5(3) - 4(0) = 15
3. C(1.5, 1.5) = 5(1.5) - 4(1.5) = 1.5
Therefore, the vertex (3, 0) maximizes the objective function.
Question 3: What is the maximum value for C?
The maximum value for C occurs at the vertex (3, 0), which is 15. Therefore, the maximum value for C is 15.
Question 1: Using a graphing calculator, find and select all the vertices for the feasible region.
To find the vertices of the feasible region, we need to find the intersection points of the following inequalities:
1) x ≥ 0
2) y ≥ 0
3) -x + 3 ≥ y
4) y ≤ (1/3)x + 1
Using the graphing calculator, we plot the inequalities and observe the points of intersection. The vertices of the feasible region are the points where the lines meet.
Verifying each option:
- (0,3): This point satisfies all the constraints.
- (3,0): This point satisfies all the constraints.
- (0,0): This point satisfies all the constraints.
- (1.5,1.5): This point does not satisfy the constraint y ≤ (1/3)x + 1.
- (-3,0): This point does not satisfy the constraint x ≥ 0.
- (1,0): This point satisfies all the constraints.
- (0,1): This point satisfies all the constraints.
Therefore, the feasible region vertices are:
(0,3), (3,0), (0,0), (1,0), and (0,1).
Question 2: Which vertex of the feasible region from above would maximize the objective function? (Show steps)
To find the vertex that maximizes the objective function, we substitute the x and y values of each vertex into the objective function C = 5x - 4y and evaluate the objective function at each vertex. The vertex with the largest objective function value will maximize the objective function.
Evaluating the objective function at each vertex:
- (0,3): C = 5(0) - 4(3) = -12
- (3,0): C = 5(3) - 4(0) = 15
- (0,0): C = 5(0) - 4(0) = 0
- (1,0): C = 5(1) - 4(0) = 5
- (0,1): C = 5(0) - 4(1) = -4
Therefore, the vertex that maximizes the objective function is (3,0).
Question 3: What is the maximum value for C?
The maximum value for C is obtained by evaluating the objective function at the vertex that maximizes it.
From Question 2, we found that the vertex (3,0) maximizes the objective function.
Substituting x = 3 and y = 0 into the objective function C = 5x - 4y:
C = 5(3) - 4(0) = 15
Therefore, the maximum value for C is 15.
To answer these questions, we need to find the feasible region and evaluate the objective function at each vertex of the region.
Question 1: Finding the vertices of the feasible region:
To graph the given constraints, we need to convert them into slope-intercept form (y = mx + b) and plot the resulting lines on a graphing calculator.
Constraint 1: -x + 3 ≥ y => y ≤ -x + 3
This constraint represents the equation of a line with slope -1 and y-intercept 3. Graph this line on a graphing calculator.
Constraint 2: y ≤ (1/3)x + 1
This constraint represents the equation of a line with slope 1/3 and y-intercept 1. Graph this line on the same graphing calculator.
Now, the feasible region is the area where both constraints are satisfied. It is the bounded region below the line y = -x + 3 and above the line y = (1/3)x + 1. The feasible region is the intersection of the shaded area below:
| /\
| /
(0,3) | * (0,1)
| /
(3,0) * | /
| (1,1)
------------------------
(3,0) (1.5,1.5)
After plotting the lines, we can see that the feasible region has vertices at the following points: (0,3), (3,0), (0,0), (1.5,1.5).
Question 2: To find the vertex that maximizes the objective function, we need to substitute the coordinates of each vertex into the objective function and choose the vertex with the maximum value.
Let's evaluate the objective function C = 5x - 4y at each vertex:
C(0,3) = 5(0) - 4(3) = -12
C(3,0) = 5(3) - 4(0) = 15
C(0,0) = 5(0) - 4(0) = 0
C(1.5,1.5) = 5(1.5) - 4(1.5) = 1.5
We can see that the vertex (3,0) maximizes the objective function C.
Question 3: The maximum value for C is determined by evaluating the objective function at the vertex that maximizes it.
From the previous evaluation, we found the maximum value at vertex (3,0), where C = 15. Therefore, the maximum value for C is 15.