Given the system of constraints, name all vertices of the feasible region. Then find the maximum value of the given objective function.constraints⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪x≥0y≥0y≤13x+35≥x+y Objective Function: C=6x−4y
To find the vertices of the feasible region, we can find the points of intersection for the lines represented by the constraints.
First, let's plot the lines on a graph:
1) x ≥ 0: This is the x-axis.
2) y ≥ 0: This is the y-axis.
3) y ≤ 13x + 35:
To plot this line, we can find the x and y intercepts.
When x = 0, y = 35.
When y = 0, 13x = -35, x ≈ -2.69.
Plotting these points, we see that the line passes through (0, 35) and approximately (-2.69, 0), trending upwards.
Next, let's find the points of intersection:
1) (0, 0): Intersection of x ≥ 0 and y ≥ 0.
2) (0, 35): Intersection of x ≥ 0 and y ≤ 13x + 35.
3) To find the third point, we need to find the intersection of y = 13x + 35 and x + y = 4.
Substituting y = 4 - x into the equation y = 13x + 35, we get:
4 - x = 13x + 35
14x = -31
x ≈ -2.21
Plugging this value of x into x + y = 4, we get y ≈ 6.21.
So the third point of intersection is approximately (-2.21, 6.21).
Now, let's evaluate the objective function at each vertex:
1) C(0, 0) = 6(0) - 4(0) = 0
2) C(0, 35) = 6(0) - 4(35) = -140
3) C(-2.21, 6.21) = 6(-2.21) - 4(6.21) ≈ -42.86
The maximum value of the objective function is -42.86, which occurs at the vertex (-2.21, 6.21).
To find the vertices of the feasible region, we need to solve the system of constraints.
1) x ≥ 0: This constraint means that x can take any non-negative value.
2) y ≥ 0: This constraint means that y can take any non-negative value.
3) y ≤ 13x + 35: This is an inequality in slope-intercept form. To find the vertex of this line, we substitute y = 13x + 35 into the inequality and solve for x:
y ≤ 13x + 35
13x + 35 ≤ 13x + 35
0 ≤ 0
Since 0 ≤ 0 is always true, this inequality is true for all values of x and y. So, this constraint does not affect the feasible region.
Now let's find the intersection of the two constraints x ≥ 0 and y ≥ 0:
x ≥ 0 and y ≥ 0
This means that both x and y must be non-negative.
Therefore, the feasible region is the entire positive quadrant (x ≥ 0, y ≥ 0), and it doesn't have any distinct vertices.
To find the maximum value of the objective function, we can evaluate the objective function at any point in the feasible region. Let's choose a point P(1, 1) from the feasible region:
C = 6x - 4y
C = 6(1) - 4(1)
C = 2
So, the maximum value of the objective function is 2. However, there is no specific vertex associated with this maximum value, since the feasible region is unbounded.
To find the vertices of the feasible region, we need to first plot the given constraints on a graph. Let's start with the constraints:
1) x ≥ 0: This constraint represents the region to the right of the y-axis.
2) y ≥ 0: This constraint represents the region above the x-axis.
3) y ≤ 13: This constraint represents a horizontal line passing through the point (0, 13).
Next, let's plot the line x + 35 ≥ x + y:
Rearranging the equation, we get:
y ≤ -35
This constraint represents a horizontal line passing through the point (-35, 0).
Plotting these constraints on a graph, the feasible region is formed by the intersection of all the shaded regions.
Now, we can identify the vertices of the feasible region by finding the points of intersection.
Since the line y ≤ 13 is parallel to the x-axis, it intersects the y-axis at (0, 13).
Since the line y ≤ -35 is parallel to the x-axis, it intersects the y-axis at (0, -35).
Since the other two constraints represent lines that extend to infinity, we need to find the intersection points with the other two constraints.
Let's solve the system of equations to find those points:
1) x + 35 = x + y
Simplifying, we get: y = 35
So, the intersection point is (-35, 35).
2) x = 0, y = 13
So, the intersection point is (0, 13).
The vertices of the feasible region are:
(0, 13), (0, -35), (-35, 35).
To find the maximum value of the objective function C = 6x - 4y, we need to evaluate it at each vertex and determine the maximum value.
Evaluating at (0, 13):
C = 6(0) - 4(13) = -52
Evaluating at (0, -35):
C = 6(0) - 4(-35) = 140
Evaluating at (-35, 35):
C = 6(-35) - 4(35) = -350
Therefore, the maximum value of the objective function is 140, which occurs at the vertex (0, -35).