Answer :
### Step-by-Step Solution
Let's break down the problem step by step to understand both the shading of the solution set on a diagram and the process of finding the maximum values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] for the given objective function [tex]\( 10x + 5y \)[/tex].
#### 1. Inequalities:
To draw the feasible region, we need to understand each inequality:
1. [tex]\( 2x - y + 4 \geq 0 \)[/tex]
2. [tex]\( x + y \leq 10 \)[/tex]
3. [tex]\( y - x \geq 0 \)[/tex]
4. [tex]\( y \geq 2 \)[/tex]
5. [tex]\( x \geq 0 \)[/tex]
#### 2. Representing Inequalities on a Diagram
Let's graph each inequality step by step:
1. Inequality [tex]\( 2x - y + 4 \geq 0 \)[/tex]:
- Rewrite as [tex]\( y \leq 2x + 4 \)[/tex].
- This is a line with slope 2 and y-intercept 4. The area below this line satisfies the inequality.
2. Inequality [tex]\( x + y \leq 10 \)[/tex]:
- This is a line with slope -1 and y-intercept 10. The area below this line satisfies the inequality.
3. Inequality [tex]\( y - x \geq 0 \)[/tex]:
- Rewrite as [tex]\( y \geq x \)[/tex].
- This is a line with a slope of 1 passing through the origin. The area above this line satisfies the inequality.
4. Inequality [tex]\( y \geq 2 \)[/tex]:
- The area above the horizontal line [tex]\( y = 2 \)[/tex] satisfies this inequality.
5. Inequality [tex]\( x \geq 0 \)[/tex]:
- The area to the right of the vertical line [tex]\( x = 0 \)[/tex] satisfies this inequality.
#### 3. Shading the Solution Region
1. Start by plotting the lines on a coordinate plane.
2. Shade the region that satisfies each inequality:
- [tex]\( y \leq 2x + 4 \)[/tex]: Shade below the line [tex]\( y = 2x + 4 \)[/tex].
- [tex]\( x + y \leq 10 \)[/tex]: Shade below the line [tex]\( x + y = 10 \)[/tex].
- [tex]\( y \geq x \)[/tex]: Shade above the line [tex]\( y = x \)[/tex].
- [tex]\( y \geq 2 \)[/tex]: Shade above the line [tex]\( y = 2 \)[/tex].
- [tex]\( x \geq 0 \)[/tex]: Shade to the right of the line [tex]\( x = 0 \)[/tex].
The feasible region is where all these shaded areas overlap.
#### 4. Finding the Vertices of the Feasible Region
Identify the points where the boundary lines intersect. These points are the vertices of the feasible region. Calculate the intersection points by solving the equations of the lines:
- Intersection of [tex]\( y = 2x + 4 \)[/tex] and [tex]\( x + y = 10 \)[/tex]:
Solve the system:
[tex]\[ \begin{cases} y = 2x + 4 \\ x + y = 10 \end{cases} \][/tex]
Substituting [tex]\( y \)[/tex]:
[tex]\[ x + 2x + 4 = 10 \implies 3x + 4 = 10 \implies x = 2, \, y = 8 \][/tex]
Intersection point: [tex]\( (2, 8) \)[/tex].
- Intersection of [tex]\( x + y = 10 \)[/tex] and [tex]\( y = 2 \)[/tex]:
Solve the system:
[tex]\[ \begin{cases} x + y = 10 \\ y = 2 \end{cases} \][/tex]
Substituting [tex]\( y \)[/tex]:
[tex]\[ x + 2 = 10 \implies x = 8 \][/tex]
Intersection point: [tex]\( (8, 2) \)[/tex].
- Intersection of [tex]\( y = 2 \)[/tex] and [tex]\( y = x \)[/tex]:
Since [tex]\( y = x \)[/tex], substituting [tex]\( y = 2 \)[/tex] gives [tex]\( x = 2 \)[/tex].
Intersection point: [tex]\( (2, 2) \)[/tex].
- Intersection of [tex]\( y = x \)[/tex] and [tex]\( y \leq 2x + 4 \)[/tex]:
Already considered in the first intersection point result: [tex]\( (2, 8) \)[/tex].
#### 5. Evaluating the Objective Function
Evaluate the objective function [tex]\( 10x + 5y \)[/tex] at each vertex:
- At [tex]\( (2, 8) \)[/tex]:
[tex]\[ 10(2) + 5(8) = 20 + 40 = 60 \][/tex]
- At [tex]\( (8, 2) \)[/tex]:
[tex]\[ 10(8) + 5(2) = 80 + 10 = 90 \][/tex]
- At [tex]\( (2, 2) \)[/tex]:
[tex]\[ 10(2) + 5(2) = 20 + 10 = 30 \][/tex]
#### 6. Conclusion
The maximum value of [tex]\( 10x + 5y \)[/tex] is obtained at the point [tex]\( (4, 4) \)[/tex], and the maximum value is 60. At this point:
[tex]\[ (x, y) = (4, 4), \quad \text{with} \quad 10x + 5y = 60 \][/tex]
Let's break down the problem step by step to understand both the shading of the solution set on a diagram and the process of finding the maximum values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] for the given objective function [tex]\( 10x + 5y \)[/tex].
#### 1. Inequalities:
To draw the feasible region, we need to understand each inequality:
1. [tex]\( 2x - y + 4 \geq 0 \)[/tex]
2. [tex]\( x + y \leq 10 \)[/tex]
3. [tex]\( y - x \geq 0 \)[/tex]
4. [tex]\( y \geq 2 \)[/tex]
5. [tex]\( x \geq 0 \)[/tex]
#### 2. Representing Inequalities on a Diagram
Let's graph each inequality step by step:
1. Inequality [tex]\( 2x - y + 4 \geq 0 \)[/tex]:
- Rewrite as [tex]\( y \leq 2x + 4 \)[/tex].
- This is a line with slope 2 and y-intercept 4. The area below this line satisfies the inequality.
2. Inequality [tex]\( x + y \leq 10 \)[/tex]:
- This is a line with slope -1 and y-intercept 10. The area below this line satisfies the inequality.
3. Inequality [tex]\( y - x \geq 0 \)[/tex]:
- Rewrite as [tex]\( y \geq x \)[/tex].
- This is a line with a slope of 1 passing through the origin. The area above this line satisfies the inequality.
4. Inequality [tex]\( y \geq 2 \)[/tex]:
- The area above the horizontal line [tex]\( y = 2 \)[/tex] satisfies this inequality.
5. Inequality [tex]\( x \geq 0 \)[/tex]:
- The area to the right of the vertical line [tex]\( x = 0 \)[/tex] satisfies this inequality.
#### 3. Shading the Solution Region
1. Start by plotting the lines on a coordinate plane.
2. Shade the region that satisfies each inequality:
- [tex]\( y \leq 2x + 4 \)[/tex]: Shade below the line [tex]\( y = 2x + 4 \)[/tex].
- [tex]\( x + y \leq 10 \)[/tex]: Shade below the line [tex]\( x + y = 10 \)[/tex].
- [tex]\( y \geq x \)[/tex]: Shade above the line [tex]\( y = x \)[/tex].
- [tex]\( y \geq 2 \)[/tex]: Shade above the line [tex]\( y = 2 \)[/tex].
- [tex]\( x \geq 0 \)[/tex]: Shade to the right of the line [tex]\( x = 0 \)[/tex].
The feasible region is where all these shaded areas overlap.
#### 4. Finding the Vertices of the Feasible Region
Identify the points where the boundary lines intersect. These points are the vertices of the feasible region. Calculate the intersection points by solving the equations of the lines:
- Intersection of [tex]\( y = 2x + 4 \)[/tex] and [tex]\( x + y = 10 \)[/tex]:
Solve the system:
[tex]\[ \begin{cases} y = 2x + 4 \\ x + y = 10 \end{cases} \][/tex]
Substituting [tex]\( y \)[/tex]:
[tex]\[ x + 2x + 4 = 10 \implies 3x + 4 = 10 \implies x = 2, \, y = 8 \][/tex]
Intersection point: [tex]\( (2, 8) \)[/tex].
- Intersection of [tex]\( x + y = 10 \)[/tex] and [tex]\( y = 2 \)[/tex]:
Solve the system:
[tex]\[ \begin{cases} x + y = 10 \\ y = 2 \end{cases} \][/tex]
Substituting [tex]\( y \)[/tex]:
[tex]\[ x + 2 = 10 \implies x = 8 \][/tex]
Intersection point: [tex]\( (8, 2) \)[/tex].
- Intersection of [tex]\( y = 2 \)[/tex] and [tex]\( y = x \)[/tex]:
Since [tex]\( y = x \)[/tex], substituting [tex]\( y = 2 \)[/tex] gives [tex]\( x = 2 \)[/tex].
Intersection point: [tex]\( (2, 2) \)[/tex].
- Intersection of [tex]\( y = x \)[/tex] and [tex]\( y \leq 2x + 4 \)[/tex]:
Already considered in the first intersection point result: [tex]\( (2, 8) \)[/tex].
#### 5. Evaluating the Objective Function
Evaluate the objective function [tex]\( 10x + 5y \)[/tex] at each vertex:
- At [tex]\( (2, 8) \)[/tex]:
[tex]\[ 10(2) + 5(8) = 20 + 40 = 60 \][/tex]
- At [tex]\( (8, 2) \)[/tex]:
[tex]\[ 10(8) + 5(2) = 80 + 10 = 90 \][/tex]
- At [tex]\( (2, 2) \)[/tex]:
[tex]\[ 10(2) + 5(2) = 20 + 10 = 30 \][/tex]
#### 6. Conclusion
The maximum value of [tex]\( 10x + 5y \)[/tex] is obtained at the point [tex]\( (4, 4) \)[/tex], and the maximum value is 60. At this point:
[tex]\[ (x, y) = (4, 4), \quad \text{with} \quad 10x + 5y = 60 \][/tex]
