Question

What point in the feasible region maximized the objective function?
{ y ≥ 0
{x ≥ 0
{-x + 3 ≥ y
{y ≤ 1/3 x + 1

Answer 1

you didn't post the objective function...

Answer 2

Oh yes! C = 5x - (-4)y

Answer 3

Oops haha

Answer 4

You need to graph each one of those inequalities... they "box" in some region... that's the "feasible region". Then you pick (x,y) points from inside that region and try those point values in your function C.

Start by graphing the four inequalities...

Answer 5

Oh yeah! I had forgotten how to do these. Thank you, I will graph them now.

Answer 6

you may have to choose a couple different points to find the maximum C, but usually, you can tell after the first one or two how to move toward the best pair for max C.

Answer 7

yes, and I forget, does y = mx + b find a line? Or does something else?

Answer 8

yes, that's a line

Answer 9

Righteo

Answer 10

The first two inequalities say that the region is to the right of the y axis and above the x axis, respectively.

Answer 11

We don't know what y is right?

Answer 12

Wait, never mind

Answer 13

Blarg I'm blanking at drawing a line. I'm sorry.

Answer 14

There, I think i'm getting it.

Answer 15

Ok, I think I might have it. I have (0, 0) and (0, 3) and (3, 2) and (3 ,0)

Answer 16

Am I close?

Answer 17

are those the corners?

Answer 18

Those were them

Answer 19

I think that sounds right... I don't have anything handy to draw it out, but I know that (0,0), (0,3), and (3,0) are right... you probably have it right :)

Answer 20

Awesome! Thank you so much!

Answer 21

I am definitely becoming a regular fan of yours! And, I apologize if I was frustrating at times.

Answer 22

Thank you!

Answer 23

I haven't done a lot of these optimization problems. Is it best to try the vertices as possible optimum points? Or are they just boundaries?

Answer 24

I think it's just bounderies

Answer 25

I know you go searching for the optimum point for C inside that region, but I've seen different things about how to best choose the points you try... it sounds like a bit of a trial and error process.

Answer 26

I think so. Well, I agree with you!

Answer 27

It seems that way! ^u^

Answer 28

Anyway, I need to be going now! Thank you so much for your help!

Answer 29

Have a wonderful day!

Answer 30

good luck :) Hope it was helpful, even though I didn't do too much other than point you in the right direction.