Answer :
To find the vertices of the feasible region for the given constraints:
[tex]\[ \begin{aligned} &2x + 3y \geq 12 \\ &5x + 2y \geq 15 \\ &x \geq 0 \\ &y \geq 0 \end{aligned} \][/tex]
we need to determine the points where the boundary lines of these inequalities intersect each other and the coordinate axes.
### Step 1: Intersection of [tex]\(2x + 3y = 12\)[/tex] and [tex]\(5x + 2y = 15\)[/tex]
To find the intersection of these lines, we solve the system of equations:
1. [tex]\(2x + 3y = 12\)[/tex]
2. [tex]\(5x + 2y = 15\)[/tex]
Solving this system, we get:
[tex]\[ x \approx 1.9090909090909092, \quad y \approx 2.727272727272727 \][/tex]
So, one vertex is [tex]\(\left(1.9090909090909092, 2.727272727272727\right)\)[/tex].
### Step 2: Intersection of [tex]\(2x + 3y = 12\)[/tex] with the x-axis
To find where [tex]\(2x + 3y = 12\)[/tex] intersects the x-axis (where [tex]\(y = 0\)[/tex]):
[tex]\[ 2x + 3(0) = 12 \implies x = 6 \][/tex]
So, another vertex is [tex]\((6, 0)\)[/tex].
### Step 3: Intersection of [tex]\(5x + 2y = 15\)[/tex] with the y-axis
To find where [tex]\(5x + 2y = 15\)[/tex] intersects the y-axis (where [tex]\(x = 0\)[/tex]):
[tex]\[ 5(0) + 2y = 15 \implies y = 7.5 \][/tex]
So, another vertex is [tex]\((0, 7.5)\)[/tex]. Note: This calculation doesn't affect our final solution because it doesn't appear in constraints.
### Step 4: Intersection of [tex]\(2x + 3y = 12\)[/tex] with the y-axis
To find where [tex]\(2x + 3y = 12\)[/tex] intersects the y-axis (where [tex]\(x = 0\)[/tex]):
[tex]\[ 2(0) + 3y = 12 \implies y = 4 \][/tex]
So, another vertex is [tex]\((0, 4)\)[/tex].
### Step 5: Intersection of [tex]\(5x + 2y = 15\)[/tex] with the x-axis
To find where [tex]\(5x + 2y = 15\)[/tex] intersects the x-axis (where [tex]\(y = 0\)[/tex]):
[tex]\[ 5x + 2(0) = 15 \implies x = 3 \][/tex]
So, another vertex is [tex]\((3, 0)\)[/tex].
Thus, the vertices of the feasible region are:
[tex]\[ (6, 0), \left(1.9090909090909092, 2.727272727272727\right), (0, 4), (3, 0) \][/tex]
These vertices match the choice:
[tex]\[ \boxed{(0,4),(1.9090909090909092,2.7272727272727273),(3,0),(6,0)} \][/tex]
[tex]\[ \begin{aligned} &2x + 3y \geq 12 \\ &5x + 2y \geq 15 \\ &x \geq 0 \\ &y \geq 0 \end{aligned} \][/tex]
we need to determine the points where the boundary lines of these inequalities intersect each other and the coordinate axes.
### Step 1: Intersection of [tex]\(2x + 3y = 12\)[/tex] and [tex]\(5x + 2y = 15\)[/tex]
To find the intersection of these lines, we solve the system of equations:
1. [tex]\(2x + 3y = 12\)[/tex]
2. [tex]\(5x + 2y = 15\)[/tex]
Solving this system, we get:
[tex]\[ x \approx 1.9090909090909092, \quad y \approx 2.727272727272727 \][/tex]
So, one vertex is [tex]\(\left(1.9090909090909092, 2.727272727272727\right)\)[/tex].
### Step 2: Intersection of [tex]\(2x + 3y = 12\)[/tex] with the x-axis
To find where [tex]\(2x + 3y = 12\)[/tex] intersects the x-axis (where [tex]\(y = 0\)[/tex]):
[tex]\[ 2x + 3(0) = 12 \implies x = 6 \][/tex]
So, another vertex is [tex]\((6, 0)\)[/tex].
### Step 3: Intersection of [tex]\(5x + 2y = 15\)[/tex] with the y-axis
To find where [tex]\(5x + 2y = 15\)[/tex] intersects the y-axis (where [tex]\(x = 0\)[/tex]):
[tex]\[ 5(0) + 2y = 15 \implies y = 7.5 \][/tex]
So, another vertex is [tex]\((0, 7.5)\)[/tex]. Note: This calculation doesn't affect our final solution because it doesn't appear in constraints.
### Step 4: Intersection of [tex]\(2x + 3y = 12\)[/tex] with the y-axis
To find where [tex]\(2x + 3y = 12\)[/tex] intersects the y-axis (where [tex]\(x = 0\)[/tex]):
[tex]\[ 2(0) + 3y = 12 \implies y = 4 \][/tex]
So, another vertex is [tex]\((0, 4)\)[/tex].
### Step 5: Intersection of [tex]\(5x + 2y = 15\)[/tex] with the x-axis
To find where [tex]\(5x + 2y = 15\)[/tex] intersects the x-axis (where [tex]\(y = 0\)[/tex]):
[tex]\[ 5x + 2(0) = 15 \implies x = 3 \][/tex]
So, another vertex is [tex]\((3, 0)\)[/tex].
Thus, the vertices of the feasible region are:
[tex]\[ (6, 0), \left(1.9090909090909092, 2.727272727272727\right), (0, 4), (3, 0) \][/tex]
These vertices match the choice:
[tex]\[ \boxed{(0,4),(1.9090909090909092,2.7272727272727273),(3,0),(6,0)} \][/tex]
