Answer :
To determine the vertices of the feasible region defined by the given constraints, we will solve the constraints pairwise to find their intersection points and then determine which of these points are feasible by checking them against all the constraints.
The given constraints are:
[tex]\[ \begin{array}{l} x + 3y \leq 6 \\ 4x + 6y \geq 9 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
### Step 1: Find Intersection Points of the Constraint Lines
Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(x = 0\)[/tex]:
[tex]\[ x = 0 \rightarrow 0 + 3y = 6 \][/tex]
[tex]\[ y = 2 \][/tex]
So, the point is (0, 2).
Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ y = 0 \rightarrow x + 0 = 6 \][/tex]
[tex]\[ x = 6 \][/tex]
So, the point is (6, 0).
Intersection of [tex]\(4x + 6y = 9\)[/tex] and [tex]\(x = 0\)[/tex]:
[tex]\[ x = 0 \rightarrow 4(0) + 6y = 9 \][/tex]
[tex]\[ 6y = 9 \][/tex]
[tex]\[ y = \frac{3}{2} \][/tex]
So, the point is (0, [tex]\(\frac{3}{2}\)[/tex]).
Intersection of [tex]\(4x + 6y = 9\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ y = 0 \rightarrow 4x + 0 = 9 \][/tex]
[tex]\[ x = \frac{9}{4} \][/tex]
So, the point is ([tex]\(\frac{9}{4}\)[/tex], 0).
Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(4x + 6y = 9\)[/tex]:
We solve this system of equations:
[tex]\[ x + 3y = 6 \][/tex]
[tex]\[ 4x + 6y = 9 \][/tex]
Multiply the first equation by 2 to align the coefficients of y:
[tex]\[ 2(x + 3y) = 2 \cdot 6 \][/tex]
[tex]\[ 2x + 6y = 12 \][/tex]
Now, subtract the second equation from this result:
[tex]\[ (2x + 6y) - (4x + 6y) = 12 - 9 \][/tex]
[tex]\[ -2x = 3 \][/tex]
[tex]\[ x = -\frac{3}{2} \][/tex]
Substitute [tex]\( x = -\frac{3}{2} \)[/tex] back into the first constraint equation:
[tex]\[ -\frac{3}{2} + 3y = 6 \][/tex]
[tex]\[ 3y = 6 + \frac{3}{2} \][/tex]
[tex]\[ 3y = \frac{12}{2} + \frac{3}{2} \][/tex]
[tex]\[ 3y = \frac{15}{2} \][/tex]
[tex]\[ y = \frac{5}{2} \][/tex]
So, the point is ([tex]\(-\frac{3}{2}, \frac{5}{2}\)[/tex]).
### Step 2: Determine Feasible Intersection Points
The potential intersection points found are:
- (0, 2)
- (6, 0)
- (0, [tex]\(\frac{3}{2}\)[/tex])
- ([tex]\(\frac{9}{4}\)[/tex], 0)
- ([tex]\(-\frac{3}{2}, \frac{5}{2}\)[/tex])
Now, we need to check each point against all constraints to determine if they lie within the feasible region:
1. (0, 2):
- [tex]\(x + 3y = 0 + 3(2) = 6 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(0) + 6(2) = 12 \geq 9\)[/tex]
- [tex]\(x = 0 \geq 0\)[/tex]
- [tex]\(y = 2 \geq 0\)[/tex]
Feasible.
2. (6, 0):
- [tex]\(x + 3y = 6 + 3(0) = 6 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(6) + 6(0) = 24 \geq 9\)[/tex]
- [tex]\(x = 6 \geq 0\)[/tex]
- [tex]\(y = 0 \geq 0\)[/tex]
Feasible.
3. (0, [tex]\(\frac{3}{2}\)[/tex]):
- [tex]\(x + 3y = 0 + 3(\frac{3}{2}) = 4.5 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(0) + 6(\frac{3}{2}) = 9 \geq 9\)[/tex]
- [tex]\(x = 0 \geq 0\)[/tex]
- [tex]\(y = \frac{3}{2} \geq 0\)[/tex]
Feasible.
4. ([tex]\(\frac{9}{4}\)[/tex], 0):
- [tex]\(x + 3y = \frac{9}{4} + 3(0) = \frac{9}{4} = 2.25 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(\frac{9}{4}) + 6(0) = 9 \geq 9\)[/tex]
- [tex]\(x = \frac{9}{4} \geq 0\)[/tex]
- [tex]\(y = 0 \geq 0\)[/tex]
Feasible.
5. ([tex]\(-\frac{3}{2}, \frac{5}{2}\)[/tex]):
- [tex]\(x + 3y = -\frac{3}{2} + 3(\frac{5}{2})\)[/tex]
- [tex]\( -\frac{3}{2} + \frac{15}{2} = \frac{12}{2} = 6 \leq 6 \)[/tex]
- [tex]\(4x + 6y = 4(-\frac{3}{2}) + 6(\frac{5}{2})\)[/tex]
- [tex]\( -6 + 15 = 9 \geq 9\)[/tex]
- [tex]\(x = -\frac{3}{2} \not\geq 0 \)[/tex]
Not feasible.
Hence, the feasible vertices of the region described by the constraints are:
[tex]\[ (0, 2), (6, 0), (0, \frac{3}{2}), (\frac{9}{4}, 0) \][/tex]
Therefore, the correct set of vertices of the feasible region is:
[tex]\[ \boxed{(0, 2), (6, 0), (0, \frac{3}{2}), (\frac{9}{4}, 0)} \][/tex]
The given constraints are:
[tex]\[ \begin{array}{l} x + 3y \leq 6 \\ 4x + 6y \geq 9 \\ x \geq 0 \\ y \geq 0 \end{array} \][/tex]
### Step 1: Find Intersection Points of the Constraint Lines
Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(x = 0\)[/tex]:
[tex]\[ x = 0 \rightarrow 0 + 3y = 6 \][/tex]
[tex]\[ y = 2 \][/tex]
So, the point is (0, 2).
Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ y = 0 \rightarrow x + 0 = 6 \][/tex]
[tex]\[ x = 6 \][/tex]
So, the point is (6, 0).
Intersection of [tex]\(4x + 6y = 9\)[/tex] and [tex]\(x = 0\)[/tex]:
[tex]\[ x = 0 \rightarrow 4(0) + 6y = 9 \][/tex]
[tex]\[ 6y = 9 \][/tex]
[tex]\[ y = \frac{3}{2} \][/tex]
So, the point is (0, [tex]\(\frac{3}{2}\)[/tex]).
Intersection of [tex]\(4x + 6y = 9\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ y = 0 \rightarrow 4x + 0 = 9 \][/tex]
[tex]\[ x = \frac{9}{4} \][/tex]
So, the point is ([tex]\(\frac{9}{4}\)[/tex], 0).
Intersection of [tex]\(x + 3y = 6\)[/tex] and [tex]\(4x + 6y = 9\)[/tex]:
We solve this system of equations:
[tex]\[ x + 3y = 6 \][/tex]
[tex]\[ 4x + 6y = 9 \][/tex]
Multiply the first equation by 2 to align the coefficients of y:
[tex]\[ 2(x + 3y) = 2 \cdot 6 \][/tex]
[tex]\[ 2x + 6y = 12 \][/tex]
Now, subtract the second equation from this result:
[tex]\[ (2x + 6y) - (4x + 6y) = 12 - 9 \][/tex]
[tex]\[ -2x = 3 \][/tex]
[tex]\[ x = -\frac{3}{2} \][/tex]
Substitute [tex]\( x = -\frac{3}{2} \)[/tex] back into the first constraint equation:
[tex]\[ -\frac{3}{2} + 3y = 6 \][/tex]
[tex]\[ 3y = 6 + \frac{3}{2} \][/tex]
[tex]\[ 3y = \frac{12}{2} + \frac{3}{2} \][/tex]
[tex]\[ 3y = \frac{15}{2} \][/tex]
[tex]\[ y = \frac{5}{2} \][/tex]
So, the point is ([tex]\(-\frac{3}{2}, \frac{5}{2}\)[/tex]).
### Step 2: Determine Feasible Intersection Points
The potential intersection points found are:
- (0, 2)
- (6, 0)
- (0, [tex]\(\frac{3}{2}\)[/tex])
- ([tex]\(\frac{9}{4}\)[/tex], 0)
- ([tex]\(-\frac{3}{2}, \frac{5}{2}\)[/tex])
Now, we need to check each point against all constraints to determine if they lie within the feasible region:
1. (0, 2):
- [tex]\(x + 3y = 0 + 3(2) = 6 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(0) + 6(2) = 12 \geq 9\)[/tex]
- [tex]\(x = 0 \geq 0\)[/tex]
- [tex]\(y = 2 \geq 0\)[/tex]
Feasible.
2. (6, 0):
- [tex]\(x + 3y = 6 + 3(0) = 6 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(6) + 6(0) = 24 \geq 9\)[/tex]
- [tex]\(x = 6 \geq 0\)[/tex]
- [tex]\(y = 0 \geq 0\)[/tex]
Feasible.
3. (0, [tex]\(\frac{3}{2}\)[/tex]):
- [tex]\(x + 3y = 0 + 3(\frac{3}{2}) = 4.5 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(0) + 6(\frac{3}{2}) = 9 \geq 9\)[/tex]
- [tex]\(x = 0 \geq 0\)[/tex]
- [tex]\(y = \frac{3}{2} \geq 0\)[/tex]
Feasible.
4. ([tex]\(\frac{9}{4}\)[/tex], 0):
- [tex]\(x + 3y = \frac{9}{4} + 3(0) = \frac{9}{4} = 2.25 \leq 6\)[/tex]
- [tex]\(4x + 6y = 4(\frac{9}{4}) + 6(0) = 9 \geq 9\)[/tex]
- [tex]\(x = \frac{9}{4} \geq 0\)[/tex]
- [tex]\(y = 0 \geq 0\)[/tex]
Feasible.
5. ([tex]\(-\frac{3}{2}, \frac{5}{2}\)[/tex]):
- [tex]\(x + 3y = -\frac{3}{2} + 3(\frac{5}{2})\)[/tex]
- [tex]\( -\frac{3}{2} + \frac{15}{2} = \frac{12}{2} = 6 \leq 6 \)[/tex]
- [tex]\(4x + 6y = 4(-\frac{3}{2}) + 6(\frac{5}{2})\)[/tex]
- [tex]\( -6 + 15 = 9 \geq 9\)[/tex]
- [tex]\(x = -\frac{3}{2} \not\geq 0 \)[/tex]
Not feasible.
Hence, the feasible vertices of the region described by the constraints are:
[tex]\[ (0, 2), (6, 0), (0, \frac{3}{2}), (\frac{9}{4}, 0) \][/tex]
Therefore, the correct set of vertices of the feasible region is:
[tex]\[ \boxed{(0, 2), (6, 0), (0, \frac{3}{2}), (\frac{9}{4}, 0)} \][/tex]
