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

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*841032

(0,120)4*0+10*1201200

(0,0)4*0+10*00

(90,0)4*90+10*0360

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

Correct option is (B)1200

Answer is 1200.

Refer below.

Step-by-step explanation:

The maximum possible value for the objective function is 1200.

d

gimmee one of those gold crowns!

~coco