Question

is anyone good with linear programming? :)

Answer 1

im decent

Answer 2

would you mind helping me?

Answer 3

Computer Programming, such as games, and web - then yes.

Answer 4

hahahah, no! linear programming :p

Answer 5

yes i just reviewed it in class

Answer 6

would you mind helping me and explaining? lol

Answer 7

Sure :)

Answer 8

alright! well it gives me this ---> {4x+3y is greater than or equal to 30
x+3y is greater than or equal to 21
x is greater than or equal to 0
y is greater than or equal to 0

Answer 9

I prefer using a graphing calculator

Answer 10

But let me try to explain

Answer 11

alright, it says maximum for c= 5x+8y

Answer 12

its a portfolio I'm doing so it says i need to rewrite the constraints in slope intercept form. idk how to do that :/

Answer 13

Okay you understand what constraints are right?

Answer 14

no.

Answer 15

Constraints are like hm how to explain this.. im going to draw it

Answer 16

is it like when you shade and all the shadings are in the same area if that makes sense?

Answer 17

so the constraints are x>(greater than or equal to) 0 and y >(greater than or equal to)0 because the graph only stayed in the 1st quadrant

Answer 18

oooh, okay!! so slope intercept form is y=mx+b correct?

Answer 19

yes it is

Answer 20

alright, can you explain to me how you would rewrite it?

Answer 21

Yes but first what is the whole question?

Answer 22

well i have a list of things to do.
1. Rewrite the constraints in slope-intercept form.
2. List all vertices of the feasible region as ordered pairs.
3. List the values of the objective function for each vertex.
4. List the maximum or minimum amount, including the x, and y-value, of the objective function.

Answer 23

Okay i know how to do that so where is the paragraph or the other part of the problem?

Answer 24

4x+3y is greater than or equal to 30
x+3y is greater than or equal to 21
x is greater than or equal to 0
y is greater than or equal to 0

minimum for
c=5x+8y
thats all i have.

Answer 25

Hm ive never done linear programming with this but i think i get it

Answer 26

alright. i truly don't understand!

Answer 27

Lol okay lets work through this together!

Answer 28

alright! thank you!

Answer 29

Welcome!
Now i believe that you got the constraints down

Answer 30

yes, those are the constraints up there! ^^^^

Answer 31

Now your objective function should be 5x+8y

Answer 32

yes. i don't know to to go about converting the constraint into slope intercept form!

Answer 33

you dont need to!

Answer 34

4x+3y >= 30

x+3y >= 21
x >= 0
y >=0
minimum for
c=5x+8y thats all i have.

Answer 35

The constraints dont have to be in slope intercept form!

Answer 36

But I would put it into standard form

Answer 37

3y >= -4x + 30
y >= -4/3*x + 30/3
y >= -4/3*x + 10

Answer 38

can you please explain how you got that!

Answer 39

If its in standard form i can teach you the rest easier

Answer 40

you start with this
4x+3y >= 30
subtract 4x from both sides

Answer 41

oh alright so we solve it as if it was an equal sign and get y by its self?

Answer 42

right

Answer 43

the only difference is, if you multiply or divide by a negative, the sign of the inequality will change (but the equality sign won't change)

Answer 44

oooh, alright!! so would that be the slope-intercept form?

Answer 45

correct, after you simplify you should get
y >= -4/3*x + 10

Answer 46

similiarly
x+3y >= 21
3y >= -x + 21
y >= -1/3*x + 7

Answer 47

then graph the 'feasible' region, that satisfies those 4 inequalities

Answer 48

alright, i think i understand those! so for those, how do i graph them?

Answer 49

you can start at the y intercepts, and then use slope

Answer 50

can i just relace y with like 1 and such?

Answer 51

Dont forget to find the vertices after graphing

Answer 52

can you explain how to graph?

Answer 53

yes when you graph you need to graph your constraints

Answer 54

you can start at the y intercept, then do rise over run to find another point

Answer 55

or you can make a table and pick arbitrary values for x

Answer 56

alrighty

Answer 57

you make a table when you are looking for the max or the min

Answer 58

here is your feasible region. I think you made a typo above in your original inequalities
http://www.wolframalpha.com/input/?i=4x%2B3y+%3C%3D+30%2C+x%2B3y+%3C%3D+21%2C+x+%3E%3D+0%2C+y+%3E%3D0+

Answer 59

well when I graphed this
"4x+3y is greater than or equal to 30
x+3y is greater than or equal to 21
x is greater than or equal to 0
y is greater than or equal to 0
minimum for c=5x+8y thats all i have."

I get this region , and notice that it is infinite
http://www.wolframalpha.com/input/?i=4x%2B3y+%3E%3D+30%2C+x%2B3y+%3E%3D+21%2C+x+%3E%3D+0%2C+y+%3E%3D0+

Answer 60

wait so if its that way, what do i do? D:

Answer 61

hmm, i guess you would still have to plug in the corners

Answer 62

you plug the vertices from the feasible region into your objective function

Answer 63

plug in the 'corners'

Answer 64

the corners here to plug into the objective function are (0,10) , (3,6), (21,0)

Answer 65

c(0,10) = 5(0) + 8(10) = 80
c(3,6) = 5(3) + 8(6) = 63
c(21,0) = 5*21 + 8*0 = 105.

so x=3, y=6 minimizes the function

Answer 66

thank you so much!! i think i figured out how to do it! thank you so much for your guys help :) <3