What point in the feasible region maximizes the objective function?
Constraints:
x≥0
y≥0
−x+3≥y
y≤ 1/3 x+1
Objective function: C = 5x – 4y
To find the point in the feasible region that maximizes the objective function C = 5x - 4y, we first need to identify the vertices of the feasible region.
1. The first constraint x ≥ 0 represents the x-axis.
2. The second constraint y ≥ 0 represents the y-axis.
3. The third constraint -x + 3 ≥ y can be rewritten as y ≤ -x + 3, which is equivalent to y = -x + 3. This line intersects the x-axis at (3,0) and the y-axis at (0,3).
4. The fourth constraint y ≤ 1/3x + 1 can be rewritten as y = 1/3x + 1. This line intersects the x-axis at (3,1) and the y-axis at (0,1).
Now we find the intersections of these four lines to determine the vertices of the feasible region. The intersections are as follows:
A: (3,0)
B: (3,1)
C: (0,3)
D: (0,1)
Now we evaluate the objective function C = 5x - 4y at each of these vertices:
A: C = 5(3) - 4(0) = 15
B: C = 5(3) - 4(1) = 11
C: C = 5(0) - 4(3) = -12
D: C = 5(0) - 4(1) = -4
The point in the feasible region that maximizes the objective function C = 5x - 4y is point A (3,0) with a value of 15.
show the steps with no explanation
1. Vertices of feasible region:
A: (3,0)
B: (3,1)
C: (0,3)
D: (0,1)
2. Evaluate objective function at each vertex:
A: C = 5(3) - 4(0) = 15
B: C = 5(3) - 4(1) = 11
C: C = 5(0) - 4(3) = -12
D: C = 5(0) - 4(1) = -4
3. Point that maximizes objective function: A (3,0) with C = 15.