Question

What is the maximum value of the feasible regions if P = 120x + 100y
Region 1 (0 ,5)
Region 2 ( 0, 4)
Region 3 (4,0)
Region 4 (3,0)

Answer 1

Answer:

The maximum value of the feasible regions is P=500

Step-by-step explanation:

we know that

To find out the maximum value of the feasible regions, substitute the value of x and the value of y of each region in the objective function P and then compare the results

we have

$P=120x+100y$

Region 1 (0,5)

For x=0, y=5

substitute

$P=120(0)+100(5)=500$

Region 2 (0,4)

For x=0, y=4

substitute

$P=120(0)+100(4)=400$

Region 3 (4,0)

For x=4, y=0

substitute

$P=120(4)+100(0)=480$

Region 4 (3,0)

For x=3, y=0

substitute

$P=120(3)+100(0)=360$

Compare

The maximum value of the feasible regions is P=500