Given the system of constraints, name all vertices of the feasible region. Then find the maximum value of the given objective function.
OBJECTIVE: G = 24 + 3B
CONSTRAINTS:
A + B ≥ 3
204 + 10B ≤ 80
B ≤ 2A
The feasible region is determined by the system of constraints.
First, let's graph the constraints to find the vertices of the feasible region.
Constraint 1: A + B ≥ 3
This equation represents a line in the coordinate plane. To graph it, we can find two points on the line:
When A = 0, B = 3
When B = 0, A = 3
Plotting these points and drawing the line through them, we have:
*[illustration 1.png]
Constraint 2: 204 + 10B ≤ 80
Simplifying this inequality, we have:
10B ≤ -124
B ≤ -12.4
Since B cannot be negative (as it represents the number of books), this constraint is not feasible.
Constraint 3: B ≤ 2A
This constraint can be graphed similarly to Constraint 1. Finding two points on the line:
When A = 0, B = 0
When A = 3, B = 6
Plotting these points and drawing the line through them, we have:
*[illustration 2.png]
The feasible region is the shaded region where the two lines intersect:
*[illustration 3.png]
Now, let's find the vertices of the feasible region:
Looking at the graph, we can see the vertices are:
(0, 3)
(3, 6)
Now, let's find the maximum value of the objective function within this feasible region.
To do this, we substitute the coordinates of each vertex into the objective function and find the maximum value.
For the vertex (0, 3):
G = 24 + 3(3)
G = 24 + 9
G = 33
For the vertex (3, 6):
G = 24 + 3(6)
G = 24 + 18
G = 42
The maximum value of the objective function G = 24 + 3B within the feasible region is 42, which occurs at the vertex (3, 6).
To find the vertices of the feasible region, we need to solve the system of constraints.
1) A + B ≥ 3
2) 204 + 10B ≤ 80
3) B ≤ 2A
Let's solve these constraints step by step:
Constraint 1: A + B ≥ 3
Rearranging this constraint, we get:
A ≥ 3 - B
Constraint 2: 204 + 10B ≤ 80
Subtracting 204 from both sides:
10B ≤ -124
Dividing both sides by 10 (and flipping the inequality):
B ≥ -12.4
Constraint 3: B ≤ 2A
Dividing both sides by 2:
A ≥ B/2
Now, let's solve these constraints graphically to find the feasible region and its vertices:
1) Plot the line A + B = 3:
To do this, we can assign some values for A and calculate B using the equation. For example, if we let A = 0, then B = 3. If we let B = 0, then A = 3. We can plot these two points and draw a line connecting them.
2) Plot the line 204 + 10B = 80:
To plot this line, we could solve for B, but instead, we can rearrange the equation to B = (80 - 204) / 10 = -12.4. This gives us a point (-12.4, 0), which we can plot.
3) Plot the line B = 2A:
We can assign some values to A and calculate B. For example, if A = 0, then B = 0. If A = 1, then B = 2. We can plot these two points and draw a line connecting them.
Now, we have three lines on the graph representing our constraints. The vertices of the feasible region are the points where the lines intersect.
To find the maximum value of the objective function G = 24 + 3B, we need to evaluate this function at each vertex and see which one gives the highest value.
Once you have drawn the lines and found the vertices, substitute the values of A and B of each vertex into the objective function G = 24 + 3B to calculate the corresponding values of G. The vertex that gives the highest value of G is the maximum value of the objective function.
To find the vertices of the feasible region, we need to solve the system of constraints. Let's start by graphing the inequalities to identify the feasible region.
1. Start by graphing the first constraint: A + B ≥ 3.
To do this, we draw a solid line with a positive slope passing through the points (0,3) and (3,0), indicating that (0,3) and (3,0) lie on the line.
2. Next, graph the second constraint: 204 + 10B ≤ 80.
To do this, we draw a solid line with a negative slope, passing through the points (7,1) and (-12,0).
3. Lastly, graph the third constraint: B ≤ 2A.
To do this, we draw a solid line with a positive slope, starting from the origin (0,0).
By visually examining the graph, we can see that the feasible region is the region where the shaded areas overlap.
To find the vertices of the feasible region, we need to find the intersection points of the lines.
The intersection points of the lines are:
- (3,0)
- (4,1)
- (7,1)
- (12,6)
Now let's find the maximum value of the objective function within the feasible region.
To maximize G = 24 + 3B, we need to substitute the values of A and B into the objective function for each vertex to find the maximum value.
1. For the point (3,0):
G = 24 + 3(0) = 24
2. For the point (4,1):
G = 24 + 3(1) = 27
3. For the point (7,1):
G = 24 + 3(1) = 27
4. For the point (12,6):
G = 24 + 3(6) = 42
Therefore, the maximum value of the objective function is G = 42, which occurs when A = 12 and B = 6.