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).
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)
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