Question

Consider the following linear programming problem: The feasible corner points are (48,84), (0,120), (0,0), (90,0). What is the maximum possible value for the objective function

Answer 1

Answer:

We have,

max 4x+10y

3x+4y<480

4x+2y<360

The feasible corner points are ( 48,84),(0,120),(0,0),(90,0)

Now, our problem is maximum type so put above feasible points in equation max 4x + 10y one by one and select one point at which our value from this equation is maximum,

(48,84) 4*48+10*84 1032

(0,120) 4*0+10*120 1200

(0,0) 4*0+10*0 0

(90,0) 4*90+10*0 360

Here we get maximum value at (0,120) which is 1200.

Correct option is (B)1200

Answer 2

Answer:

Answer is 1200.

Refer below.

Step-by-step explanation:

The maximum possible value for the objective function is 1200.