Question

Pleassssssssssse help!
I only need help with number 4!

List all vertices of the feasible region as ordered pairs.
List the values of the objective function for each vertex.
http://i45.tinypic.com/2entf6t.png%201-4

Answer 1

First, I am suspicious that we are missing hte usual "y > 0"
Second, how shall we proceed? Are we graphing?

Answer 2

Yeah graphing.

Answer 3

Did you solve the three constraints for 'y' and identify the feasible region?

Answer 4

I don't know what to do because there are three of them instead of two.

Answer 5

I think the y intercept forms are
• Y ≥ -x/2+12
• Y ≥ -x/5+4
• Y ≥ -3x+29
• X ≥ 0

Answer 6

Try the middle one again.

Answer 7

I guess I don't know what you mean cause the original thing was x+5y≥20
and I subtract x from both sides then divided by 5 and even when I try again I got the same equation..

Answer 8

?? That appears to be the middle one from Problem #3. Are we looking at the same problems?

Answer 9

oh crap. oops my bad.
The Y would be Y≥-3x/2+17

Answer 10

and actually I just realized the signs would be ≤

Answer 11

No matter about the signs for now. We just need the lines. How good are you at drawing these lines? Is the Slope-Intercept form good enough for your fluent usage or shall we try something else?

Answer 12

I can try I don't know how good I'll be.

Answer 13

Well, you don't HAVE to graph them. That just helps visualize what you are doing. If you want, you can just solve the three systems created by pairing off the three Equalities two at a time.

Answer 14

yeah I can't.........

Answer 15

I know I have to graph them,. I need help graphing them cause i can't.

Answer 16

That might be close. Make sure all the slopes are negative and it might be better.

Answer 17

Want to know an easier way to graph such things?

Answer 18

Yeah that'd help.

Answer 19

Have you EVER heard of the "Intercept Form"? Most folks have not.

Answer 20

I did Y intercept form?

Answer 21

No, that was the standard Slope-Intercept.

The intercept form is very easy if it is what you need. In this cae, it might be. It's like this: \[\frac{ x }{ a } + \frac{ y }{ b } = 1\] The x-intercept is (a,0) and the y-intercept is (0,b). Find those two, and you're done. This is particularly applicable in your linear programming, since we are interested only in the first wuadrant.

Answer 22

okay?

Answer 23

First one: x + 2y = 24
Divide by 24 and we're done: \[\frac{x}{24} + \frac{y}{12} = 1\]

Answer 24

so that vertice group thing would be (0,1)?

Answer 25

No, no... x-intercept is (24,0) y-intercept is (0,12)

Answer 26

Second one is a little trickier. 3x + 2y = 34
Divide by 34 \[\frac{3x}{34} + \frac{2y}{34} = 1\]
Rewite just a little: \[\frac{x}{34/3} + \frac{y}{17} = 1\]

x-intercept is (34/3,0)
y-intercept is (0,17)

Answer 27

this doesn't even make sense.

Answer 28

That's a little disappointing, because it is kind of magic if you can get it. You will be the fastest 1st Quadrant grapher in your class!

Answer 29

how does this help me find the vertices?

Answer 30

If you can graph them, you can find them.

Answer 31

No worries. We can just work with the Slope-Intercept and NOT graph them. I would like to encourage you to get better at graphing. You can just keep the Intercept Form in teh back of your brain, somewhere. It will be okay hiding until youare ready for it. :-)

Answer 32

So, solve the first two: -x/2+12 = -3x/2+17

Answer 33

I still don't understand..
what am I solving for?
I thought I already solved

Answer 34

You put the constraints in slope-intercept form so that it woudl be easier to graph them. Now, the challenge is to see where the constraints intersect. Solving that equation I just showed will provide the x-coordinate of the intersection of two of teh contracts, #1 and #2.

Answer 35

12-17
-3x/2 - -x/2
?

Answer 36

Keep it up until you get x = 5.

Answer 37

it'd be -5
and the bottom would be 1?

Answer 38

Can't be -5. It has to be in the 1st Quadrant.

Answer 39

Somehow, we have to solve this system of equations:

x + 2y = 24
3x + 2y = 34

How are we going to do that?

Answer 40

You're confusing me so much.

Answer 41

Truthfully, if you can't graph it and you can't solve it, where can we go? We have to do one or the other. We can get through either one. You pick.

Answer 42

Did you solve the first three problems? How did you do those?

Answer 43

I don't need help with the other ones though, I did those by using my book but I don't understand how to do the last one because there are three equations and I've tried numerous times to solve it but I can't.

Answer 44

I see. Well, then here's a plan. Let's just ignore one equation and play like it doesn't exist. This is will be exactly like the others. We'll bring it back later, don't worry.

What do you think?

Answer 45

I don't know my brain is fried now.

Answer 46

Fair enough. Shall we beat on your fired brain a little more or would you rather come back when you are ready for another go. Give that idea about just ignoring one constraint a little thought, but not so much thought taht you give up in it. You'll get it.

Answer 47

well I need it like right now so yeah.. we can try again.. thanks

Answer 48

Using Mathematica for the calculation,\[\text{Maximize}[\{2x+3y,x+2y\leq 24,3x+2y\leq 34,3x+y\leq 29,x\geq 0\},\{x,y\}] \]\[\left\{\frac{77}{2},\left\{x=5,y=\frac{19}{2}\right\}\right\} \]Maximum function value is 77/2 at point (5,19/2)