Question

X>0 y>0-x+3>y y<1/3x+1
Use the following constraints and objective function to answer the next 3 questions. Objective function: C=5x−4y Question 1 Using a graphing calculator, find and select all the vertices for the feasible region.(1 point) Responses (0,3) (0,3) (1,0) (1,0) (-3,0) (-3,0) (3,0) (3,0) (1.5,1.5) (1.5,1.5) (0,1) (0,1) (0,0)

Answer 1

To find the vertices of the feasible region, we need to find the points where the lines representing the constraints intersect.

First, let's find the intersection of the lines y = 0 - x + 3 and y = 1/3x + 1:

0 - x + 3 = 1/3x + 1

Multiply both sides of the equation by 3 to eliminate the fraction:

0 - 3x + 9 = x + 3

Add 3x to both sides:

9 = x + 3x + 3

Combine like terms:

9 = 4x + 3

Subtract 3 from both sides:

6 = 4x

Divide by 4:

x = 3/2 = 1.5

Now let's substitute this value of x into one of the equations to find y:

y = 0 - x + 3 = 0 - 1.5 + 3 = 1.5

So the first vertex is (1.5, 1.5).

Next, let's find the intersection of the lines y = 0 - x + 3 and y = 0, with x > 0:

0 - x + 3 = 0

Solve for x:

x = 3

Substitute this value of x into y = 0 - x + 3:

y = 0 - 3 + 3 = 0

So the second vertex is (3, 0).

Finally, let's find the intersection of the lines y = 1/3x + 1 and y = 0, with x > 0:

1/3x + 1 = 0

Subtract 1 from both sides:

1/3x = -1

Multiply both sides by 3 to eliminate the fraction:

x = -3

Substitute this value of x into y = 1/3x + 1:

y = 1/3(-3) + 1 = -1 + 1 = 0

So the third vertex is (-3, 0).

Therefore, the vertices of the feasible region are (1.5, 1.5), (3, 0), and (-3, 0).