The vertices of a feasible region are (2,2) (4,4) (6,2) (6,0) (2,0). Find the maximum value of the function: P = 5x + y

Question

The vertices of a feasible region are (2,2) (4,4) (6,2) (6,0) (2,0). Find the maximum value of the function: P = 5x + y

Answer 1

To find the maximum value of P = 5x + y subject to the constraints represented by the feasible region, we need to evaluate the objective function at each of the vertices of the feasible region and determine which vertex gives the highest value of P:

P(2,2) = 5(2) + 2 = 12

P(4,4) = 5(4) + 4 = 24

P(6,2) = 5(6) + 2 = 32

P(6,0) = 5(6) + 0 = 30

P(2,0) = 5(2) + 0 = 10

Therefore, the maximum value of P = 32, occurs at the vertex (6,2), which is our maximum!

Answer 2

Substitute each pair of vlaues to determine the maximum value of P.

Answer 3

To find the maximum value of the function P = 5x + y, we need to evaluate P at each vertex of the feasible region and determine which vertex gives the highest value of P.

Vertex 1: (2,2)

P = 5x + y = 5(2) + 2 = 12

Vertex 2: (4,4)

P = 5x + y = 5(4) + 4 = 24

Vertex 3: (6,2)

P = 5x + y = 5(6) + 2 = 32

Vertex 4: (6,0)

P = 5x + y = 5(6) + 0 = 30

Vertex 5: (2,0)

P = 5x + y = 5(2) + 0 = 10

Therefore, the maximum value of the function P = 5x + y is 32, which occurs at the vertex (6,2) of the feasible region.