Determine the 5 vertices of the feasible region, given the constraints below.
3x+5y≤30
x≤9
y≤3
x≥0
y≥0
The feasible region represents the set of points that satisfy all the given constraints.
1. To find the vertex where 3x + 5y = 30 intersects with x = 0, we set x = 0 and solve for y:
3(0) + 5y = 30
5y = 30
y = 6
So the first vertex is (0, 6).
2. To find the vertex where 3x + 5y = 30 intersects with y = 0, we set y = 0 and solve for x:
3x + 5(0) = 30
3x = 30
x = 10
However, x should be less than or equal to 9 according to the constraint x ≤ 9. So this point does not fall within the feasible region.
3. To find the vertex where x = 9 intersects with the line y = 0, we set x = 9 and y = 0:
3(9) + 5(0) = 27
So the second vertex is (9, 0).
4. To find the vertex where x = 9 intersects with the line y = 3, we set x = 9 and y = 3:
3(9) + 5(3) = 27 + 15 = 42
However, 42 is greater than 30 according to the constraint 3x + 5y ≤ 30. So this point does not fall within the feasible region.
5. To find the vertex where y = 3 intersects with the line x = 0, we set x = 0 and y = 3:
3(0) + 5(3) = 0 + 15 = 15
However, 15 is greater than 9 according to the constraint x ≤ 9. So this point does not fall within the feasible region.
Thus, the feasible region is bounded by the vertices (0, 6) and (9, 0).
Note: The constraints x ≥ 0 and y ≥ 0 indicate that the feasible region must be in the first quadrant, so the feasible region is the triangle formed by the points (0, 0), (0, 6), and (9, 0).
To determine the 5 vertices of the feasible region, we can graph the given constraints and find the intersection points. Let's start by graphing the constraints one by one.
1. 3x + 5y ≤ 30:
To graph this constraint, we first need to turn it into an equation. Let's set it equal to 30:
3x + 5y = 30
Now we can solve for y:
5y = 30 - 3x
y = (30 - 3x) / 5
Now, let's choose some x-values and find the corresponding y-values to plot the points:
For x = 0: y = (30 - 3(0)) / 5 = 6 / 5 = 1.2
So, one point is (0, 1.2).
For x = 5: y = (30 - 3(5)) / 5 = 15 / 5 = 3
So, another point is (5, 3).
Plot these two points on the graph.
2. x ≤ 9:
This constraint is a vertical line at x = 9. Plot this line on the graph.
3. y ≤ 3:
This constraint is a horizontal line at y = 3. Plot this line on the graph.
4. x ≥ 0:
This constraint is a vertical line at x = 0 (y-axis). Plot this line on the graph.
5. y ≥ 0:
This constraint is a horizontal line at y = 0 (x-axis). Plot this line on the graph.
Now, let's find the intersection points of the graphed lines.
- The intersection points between constraint 1 and constraint 2 are (5, 3) and (9, 3).
- The intersection points between constraint 1 and constraint 3 are (0, 1.2) and (0, 3).
- The intersection points between constraint 2 and constraint 5 are (0, 0) and (9, 0).
- The intersection point between constraint 3 and constraint 5 is (0, 3).
- The intersection points between constraint 4 and constraint 5 are (0, 0) and (0, 3).
So, the 5 vertices of the feasible region are:
1. (0, 0)
2. (0, 1.2)
3. (0, 3)
4. (5, 3)
5. (9, 3)
To determine the vertices of the feasible region, we need to find the points where the boundary lines of the constraints intersect.
Given the constraints:
1. 3x + 5y ≤ 30
2. x ≤ 9
3. y ≤ 3
4. x ≥ 0
5. y ≥ 0
Let's solve the inequalities one by one:
1. From constraint 4 (x ≥ 0), we know that the x-coordinate of any point in the feasible region cannot be negative.
2. From constraint 5 (y ≥ 0), we know that the y-coordinate of any point in the feasible region cannot be negative.
3. For constraint 2 (x ≤ 9), the boundary line is a vertical line passing through x = 9.
4. For constraint 3 (y ≤ 3), the boundary line is a horizontal line passing through y = 3.
5. For constraint 1 (3x + 5y ≤ 30), we can rewrite it in terms of x and y:
3x + 5y = 30
5y = 30 - 3x
y = (30 - 3x) / 5
Now we can find the vertices by looking at the intersections of the boundary lines:
1. The x-intercept of the line from constraint 1 is found when y = 0:
5(0) = 30 - 3x
x = 10
Vertex: (10, 0)
2. The intersection of the lines from constraints 1 and 3 is found by solving the system of equations:
3x + 5y = 30
y = 3
Substitute y = 3 in the equation:
3x + 5(3) = 30
3x + 15 = 30
3x = 15
x = 5
Vertex: (5, 3)
3. The intersection of the lines from constraints 3 and 4 is found at the origin (0, 0).
4. The intersection of the lines from constraints 2 and 4 is found when x = 0:
y ≤ 3
y = 3
Vertex: (0, 3)
5. Finally, the intersection of the lines from constraints 2 and 3 is found at the point where y = 3 and x = 9 (from constraint 2).
Vertex: (9, 3)
Therefore, the 5 vertices of the feasible region are:
(10, 0), (5, 3), (0, 0), (0, 3), and (9, 3).