Answer :
Sure! To solve this system of inequalities graphically, we need to follow a systematic approach. Let's break it down step-by-step:
1. Convert each inequality into an equation. These equations will form the boundaries of the feasible region.
[tex]\[ 1. \quad -6x - 3y = -48 \quad \Rightarrow \quad 6x + 3y = 48 \][/tex]
[tex]\[ 2. \quad -x + 2y = 7 \][/tex]
[tex]\[ 3. \quad 7x + 6y = 91 \][/tex]
[tex]\[ 4. \quad x = 0 \][/tex]
[tex]\[ 5. \quad y = 0 \][/tex]
2. Graph each equation.
- For [tex]\(6x + 3y = 48\)[/tex], find the intercepts.
- [tex]\(x\)[/tex]-intercept: Set [tex]\(y = 0\)[/tex], then [tex]\(6x = 48 \Rightarrow x = 8\)[/tex].
- [tex]\(y\)[/tex]-intercept: Set [tex]\(x = 0\)[/tex], then [tex]\(3y = 48 \Rightarrow y = 16\)[/tex].
- For [tex]\(-x + 2y = 7\)[/tex], find the intercepts.
- [tex]\(x\)[/tex]-intercept: Set [tex]\(y = 0\)[/tex], then [tex]\(-x = 7 \Rightarrow x = -7\)[/tex]. This is outside the positive quadrant and not relevant to our graph.
- [tex]\(y\)[/tex]-intercept: Set [tex]\(x = 0\)[/tex], then [tex]\(2y = 7 \Rightarrow y = 3.5\)[/tex].
- For [tex]\(7x + 6y = 91\)[/tex], find the intercepts.
- [tex]\(x\)[/tex]-intercept: Set [tex]\(y = 0\)[/tex], then [tex]\(7x = 91 \Rightarrow x = 13\)[/tex].
- [tex]\(y\)[/tex]-intercept: Set [tex]\(x = 0\)[/tex], then [tex]\(6y = 91 \Rightarrow y = 15.17\)[/tex].
- For [tex]\(x = 0\)[/tex], this is the y-axis.
- For [tex]\(y = 0\)[/tex], this is the x-axis.
3. Shade the feasible region. Remember to shade the region that satisfies the inequalities:
- For the first inequality [tex]\(6x + 3y \leq 48\)[/tex], shade below the line.
- For the second inequality [tex]\(-x + 2y \leq 7\)[/tex], shade below the line.
- For the third inequality [tex]\(7x + 6y \leq 91\)[/tex], shade below the line.
- For [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex], shade the first quadrant.
4. Identify the vertices of the feasible region. The vertices are the points where the boundary lines intersect within the first quadrant. Solve for the points of intersection:
- Intersection of [tex]\(6x + 3y = 48\)[/tex] and [tex]\(-x + 2y = 7\)[/tex]:
- Solve [tex]\(6x + 3y = 48\)[/tex] for [tex]\(y\)[/tex]:
[tex]\[ 3y = 48 - 6x \quad \Rightarrow \quad y = \frac{48 - 6x}{3} = 16 - 2x \][/tex]
- Substitute [tex]\(y = 16 - 2x\)[/tex] into [tex]\(-x + 2(16 - 2x) = 7\)[/tex]:
[tex]\[ -x + 32 - 4x = 7 \quad \Rightarrow \quad -5x + 32 = 7 \quad \Rightarrow \quad -5x = -25 \quad \Rightarrow \quad x = 5 \quad \Rightarrow \quad y = 16 - 2(5) = 6 \][/tex]
- Intersection at [tex]\((5, 6)\)[/tex].
- Intersection of [tex]\(7x + 6y = 91\)[/tex] and [tex]\(-x + 2y = 7\)[/tex]:
- Solve [tex]\(-x + 2y = 7\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[ x = 2y - 7 \][/tex]
- Substitute [tex]\(x = 2y - 7\)[/tex] into [tex]\(7(2y - 7) + 6y = 91\)[/tex]:
[tex]\[ 14y - 49 + 6y = 91 \quad \Rightarrow \quad 20y - 49 = 91 \quad \Rightarrow \quad 20y = 140 \quad \Rightarrow \quad y = 7 \quad \Rightarrow \quad x = 2(7) - 7 = 7 \][/tex]
- Intersection at [tex]\((7, 7)\)[/tex].
- Intersection of [tex]\(6x + 3y = 48\)[/tex] and [tex]\(7x + 6y = 91\)[/tex]:
- Multiply [tex]\(6x + 3y = 48\)[/tex] by 2:
[tex]\[ 12x + 6y = 96 \][/tex]
- Subtract [tex]\(7x + 6y = 91\)[/tex] from [tex]\(12x + 6y = 96\)[/tex]:
[tex]\[ 12x + 6y - 7x - 6y = 96 - 91 \quad \Rightarrow \quad 5x = 5 \quad \Rightarrow \quad x = 1 \quad \Rightarrow \quad y = 16 - 2(1) = 14 \][/tex]
- Intersection at [tex]\((1, 14)\)[/tex].
5. List the vertices of the feasible region:
[tex]\[ \{(0, 0), (5, 6), (7, 7), (1, 14)\} \][/tex]
The feasible region is bounded by these vertices in the first quadrant.
1. Convert each inequality into an equation. These equations will form the boundaries of the feasible region.
[tex]\[ 1. \quad -6x - 3y = -48 \quad \Rightarrow \quad 6x + 3y = 48 \][/tex]
[tex]\[ 2. \quad -x + 2y = 7 \][/tex]
[tex]\[ 3. \quad 7x + 6y = 91 \][/tex]
[tex]\[ 4. \quad x = 0 \][/tex]
[tex]\[ 5. \quad y = 0 \][/tex]
2. Graph each equation.
- For [tex]\(6x + 3y = 48\)[/tex], find the intercepts.
- [tex]\(x\)[/tex]-intercept: Set [tex]\(y = 0\)[/tex], then [tex]\(6x = 48 \Rightarrow x = 8\)[/tex].
- [tex]\(y\)[/tex]-intercept: Set [tex]\(x = 0\)[/tex], then [tex]\(3y = 48 \Rightarrow y = 16\)[/tex].
- For [tex]\(-x + 2y = 7\)[/tex], find the intercepts.
- [tex]\(x\)[/tex]-intercept: Set [tex]\(y = 0\)[/tex], then [tex]\(-x = 7 \Rightarrow x = -7\)[/tex]. This is outside the positive quadrant and not relevant to our graph.
- [tex]\(y\)[/tex]-intercept: Set [tex]\(x = 0\)[/tex], then [tex]\(2y = 7 \Rightarrow y = 3.5\)[/tex].
- For [tex]\(7x + 6y = 91\)[/tex], find the intercepts.
- [tex]\(x\)[/tex]-intercept: Set [tex]\(y = 0\)[/tex], then [tex]\(7x = 91 \Rightarrow x = 13\)[/tex].
- [tex]\(y\)[/tex]-intercept: Set [tex]\(x = 0\)[/tex], then [tex]\(6y = 91 \Rightarrow y = 15.17\)[/tex].
- For [tex]\(x = 0\)[/tex], this is the y-axis.
- For [tex]\(y = 0\)[/tex], this is the x-axis.
3. Shade the feasible region. Remember to shade the region that satisfies the inequalities:
- For the first inequality [tex]\(6x + 3y \leq 48\)[/tex], shade below the line.
- For the second inequality [tex]\(-x + 2y \leq 7\)[/tex], shade below the line.
- For the third inequality [tex]\(7x + 6y \leq 91\)[/tex], shade below the line.
- For [tex]\(x \geq 0\)[/tex] and [tex]\(y \geq 0\)[/tex], shade the first quadrant.
4. Identify the vertices of the feasible region. The vertices are the points where the boundary lines intersect within the first quadrant. Solve for the points of intersection:
- Intersection of [tex]\(6x + 3y = 48\)[/tex] and [tex]\(-x + 2y = 7\)[/tex]:
- Solve [tex]\(6x + 3y = 48\)[/tex] for [tex]\(y\)[/tex]:
[tex]\[ 3y = 48 - 6x \quad \Rightarrow \quad y = \frac{48 - 6x}{3} = 16 - 2x \][/tex]
- Substitute [tex]\(y = 16 - 2x\)[/tex] into [tex]\(-x + 2(16 - 2x) = 7\)[/tex]:
[tex]\[ -x + 32 - 4x = 7 \quad \Rightarrow \quad -5x + 32 = 7 \quad \Rightarrow \quad -5x = -25 \quad \Rightarrow \quad x = 5 \quad \Rightarrow \quad y = 16 - 2(5) = 6 \][/tex]
- Intersection at [tex]\((5, 6)\)[/tex].
- Intersection of [tex]\(7x + 6y = 91\)[/tex] and [tex]\(-x + 2y = 7\)[/tex]:
- Solve [tex]\(-x + 2y = 7\)[/tex] for [tex]\(x\)[/tex]:
[tex]\[ x = 2y - 7 \][/tex]
- Substitute [tex]\(x = 2y - 7\)[/tex] into [tex]\(7(2y - 7) + 6y = 91\)[/tex]:
[tex]\[ 14y - 49 + 6y = 91 \quad \Rightarrow \quad 20y - 49 = 91 \quad \Rightarrow \quad 20y = 140 \quad \Rightarrow \quad y = 7 \quad \Rightarrow \quad x = 2(7) - 7 = 7 \][/tex]
- Intersection at [tex]\((7, 7)\)[/tex].
- Intersection of [tex]\(6x + 3y = 48\)[/tex] and [tex]\(7x + 6y = 91\)[/tex]:
- Multiply [tex]\(6x + 3y = 48\)[/tex] by 2:
[tex]\[ 12x + 6y = 96 \][/tex]
- Subtract [tex]\(7x + 6y = 91\)[/tex] from [tex]\(12x + 6y = 96\)[/tex]:
[tex]\[ 12x + 6y - 7x - 6y = 96 - 91 \quad \Rightarrow \quad 5x = 5 \quad \Rightarrow \quad x = 1 \quad \Rightarrow \quad y = 16 - 2(1) = 14 \][/tex]
- Intersection at [tex]\((1, 14)\)[/tex].
5. List the vertices of the feasible region:
[tex]\[ \{(0, 0), (5, 6), (7, 7), (1, 14)\} \][/tex]
The feasible region is bounded by these vertices in the first quadrant.
