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To graph the system of constraints and find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that maximize the objective function [tex]\( C = 7x - 3y \)[/tex], we will follow these steps:
### Step 1: Graph the Constraints
1. Constraint 1: [tex]\( x \geq 0 \)[/tex]
- This means all feasible points are on or to the right of the y-axis.
2. Constraint 2: [tex]\( y \geq 0 \)[/tex]
- This means all feasible points are on or above the x-axis.
3. Constraint 3: [tex]\( y \leq \frac{1}{5} x + 2 \)[/tex]
- To graph this, start with the intercepts. When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex], so we have the point (0, 2).
- When [tex]\( y = 0 \)[/tex], solve [tex]\( 0 = \frac{1}{5} x + 2 \)[/tex] to get [tex]\( x = -10 \)[/tex]. However, since [tex]\( x \geq 0 \)[/tex], we do not need this point so we get the intercept at (10, 0).
4. Constraint 4: [tex]\( y + x \leq 5 \)[/tex]
- To graph this, start with the intercepts again. When [tex]\( x = 0 \)[/tex], [tex]\( y = 5 \)[/tex], so we have the point (0, 5).
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 5 \)[/tex], so we have the point (5, 0).
### Step 2: Identify the Feasible Region
- The feasible region is the area where all constraints overlap.
- It will be bounded by the lines [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y + x = 5 \)[/tex] along with the lines [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex].
### Step 3: Find the Corner Points
- We need to find intersection points of the boundary lines, as the maximum value will occur at one of the vertices of the feasible region.
1. Intersection of [tex]\( y = 0 \)[/tex] and [tex]\( x = 0 \)[/tex]
- The origin (0, 0).
2. Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y + x = 5 \)[/tex]
- Solve:
[tex]\( \frac{1}{5}x + 2 + x = 5 \)[/tex]
[tex]\( \rightarrow \frac{1}{5}x + x = 3 \)[/tex]
[tex]\( \rightarrow 1.2x = 3 \)[/tex]
[tex]\( \rightarrow x = 2.5 \)[/tex]
Substitute [tex]\( x = 2.5 \)[/tex] into [tex]\( y = \frac{1}{5}x + 2 \)[/tex]:
[tex]\( y = \frac{1}{5}(2.5) + 2 = 0.5 + 2 = 2.5 \)[/tex]
- So, the point is (2.5, 2.5).
3. Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y = 0 \)[/tex]
- Already found as (0, 2) from constraints.
4. Intersection of [tex]\( y + x = 5 \)[/tex] with x and y bounds
- Points are (5, 0) and (0, 5).
### Step 4: Evaluate Objective Function at Each Vertex
1. At (0, 0):
[tex]\( C = 7(0) - 3(0) = 0 \)[/tex]
2. At (0, 2):
[tex]\( C = 7(0) - 3(2) = -6 \)[/tex]
3. At (2.5, 2.5):
[tex]\( C = 7(2.5) - 3(2.5) = 17.5 - 7.5 = 10 \)[/tex]
4. At (5, 0):
[tex]\( C = 7(5) - 3(0) = 35 \)[/tex]
5. At (0, 5):
Points outside boundary, so not evaluated.
### Conclusion
The maximum value of the objective function [tex]\( C \)[/tex] occurs at [tex]\( (2.5, 2.5) \)[/tex] and is given by:
[tex]\[ C = 10 \][/tex]
Thus, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that maximize the objective function [tex]\( C = 7x - 3y \)[/tex] are [tex]\( \boxed{(2.5, 2.5)} \)[/tex].
### Step 1: Graph the Constraints
1. Constraint 1: [tex]\( x \geq 0 \)[/tex]
- This means all feasible points are on or to the right of the y-axis.
2. Constraint 2: [tex]\( y \geq 0 \)[/tex]
- This means all feasible points are on or above the x-axis.
3. Constraint 3: [tex]\( y \leq \frac{1}{5} x + 2 \)[/tex]
- To graph this, start with the intercepts. When [tex]\( x = 0 \)[/tex], [tex]\( y = 2 \)[/tex], so we have the point (0, 2).
- When [tex]\( y = 0 \)[/tex], solve [tex]\( 0 = \frac{1}{5} x + 2 \)[/tex] to get [tex]\( x = -10 \)[/tex]. However, since [tex]\( x \geq 0 \)[/tex], we do not need this point so we get the intercept at (10, 0).
4. Constraint 4: [tex]\( y + x \leq 5 \)[/tex]
- To graph this, start with the intercepts again. When [tex]\( x = 0 \)[/tex], [tex]\( y = 5 \)[/tex], so we have the point (0, 5).
- When [tex]\( y = 0 \)[/tex], [tex]\( x = 5 \)[/tex], so we have the point (5, 0).
### Step 2: Identify the Feasible Region
- The feasible region is the area where all constraints overlap.
- It will be bounded by the lines [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y + x = 5 \)[/tex] along with the lines [tex]\( x = 0 \)[/tex] and [tex]\( y = 0 \)[/tex].
### Step 3: Find the Corner Points
- We need to find intersection points of the boundary lines, as the maximum value will occur at one of the vertices of the feasible region.
1. Intersection of [tex]\( y = 0 \)[/tex] and [tex]\( x = 0 \)[/tex]
- The origin (0, 0).
2. Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y + x = 5 \)[/tex]
- Solve:
[tex]\( \frac{1}{5}x + 2 + x = 5 \)[/tex]
[tex]\( \rightarrow \frac{1}{5}x + x = 3 \)[/tex]
[tex]\( \rightarrow 1.2x = 3 \)[/tex]
[tex]\( \rightarrow x = 2.5 \)[/tex]
Substitute [tex]\( x = 2.5 \)[/tex] into [tex]\( y = \frac{1}{5}x + 2 \)[/tex]:
[tex]\( y = \frac{1}{5}(2.5) + 2 = 0.5 + 2 = 2.5 \)[/tex]
- So, the point is (2.5, 2.5).
3. Intersection of [tex]\( y = \frac{1}{5}x + 2 \)[/tex] and [tex]\( y = 0 \)[/tex]
- Already found as (0, 2) from constraints.
4. Intersection of [tex]\( y + x = 5 \)[/tex] with x and y bounds
- Points are (5, 0) and (0, 5).
### Step 4: Evaluate Objective Function at Each Vertex
1. At (0, 0):
[tex]\( C = 7(0) - 3(0) = 0 \)[/tex]
2. At (0, 2):
[tex]\( C = 7(0) - 3(2) = -6 \)[/tex]
3. At (2.5, 2.5):
[tex]\( C = 7(2.5) - 3(2.5) = 17.5 - 7.5 = 10 \)[/tex]
4. At (5, 0):
[tex]\( C = 7(5) - 3(0) = 35 \)[/tex]
5. At (0, 5):
Points outside boundary, so not evaluated.
### Conclusion
The maximum value of the objective function [tex]\( C \)[/tex] occurs at [tex]\( (2.5, 2.5) \)[/tex] and is given by:
[tex]\[ C = 10 \][/tex]
Thus, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that maximize the objective function [tex]\( C = 7x - 3y \)[/tex] are [tex]\( \boxed{(2.5, 2.5)} \)[/tex].
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