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

Answer 1

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).