Question

Using Lagrange multipliers to find the maxium and minimum values of the function f(x,y,z)+xy^2z^3 subject to x^2+y^2+2z^2=25. PLEASE HELP

Answer 1

So you're gonna go through the equation taking derivatives with respect to all three variables but one at a time. So first, derive the equation with respect to the variable 'x' while treating y,z like constants, that's one derivative. Then do the same again but with only y as the variable, and then again with z.

Answer 2

Let me know when you have done that much!

Answer 3

so I need to take partials in respect to each variable fx, fy, fz?

Answer 4

so fx=y^2z^3
fy=2xyz^3
fz=3xy^2z^2

Answer 5

yes, partials. gimme a second

Answer 6

wait a second, rephrase your question. what do you mean by f(x,y,z)+xy^2z^3 ??

Answer 7

do you have two functions?

Answer 8

f(x,y,z)=xy^2z^3 subject to the constraint of x^2+y^2+2z^2=25

Answer 9

I'm in the middle of two other quesitons. If you're not rushed, I can get back to you on this one. alright?

Answer 10

not rushed it is due in the morning, thank you for your help. this site is fun

Answer 11

brb

Answer 12

well, in 30

Answer 13

okay

Answer 14

i couldn't get thru, som' about about the server not receiving my post but I'm still looking at the problem. Just grabbed my snack and I'm not browsing my calc book.

Answer 15

thanks

Answer 16

it's an optimization. very much in the line of those classic optimization problems in simpler calculus but I haven't done one in a bit

Answer 17

made any progress?

Answer 18

i am confused because i left my book at home and I am at school

Answer 19

something to do with taking the gradient of the constraint

Answer 20

we have not done this type of problem, and i have no information in my notes

Answer 21

which calculus are you taking?

Answer 22

calc 4, 3-d vectors, multivarible

Answer 23

nice

Answer 24

problem is only worth 5 points but i think it is important to know how to do it

Answer 25

i agree

Answer 26

i'm dining and browsing pages at the same time...

Answer 27

in a weird way, i miss school; funny how that works

Answer 28

eat, i dont want to ruin your dinner, just get back to me when you are finished

Answer 29

don't worry about it

Answer 30

i only have 2 more math classes and 3 physical chem classes and 3 chem engineering classes and then I am done

Answer 31

lol, then heading for the petroleum industry?

Answer 32

i want to do desalinization, water filtration, water is the next oil!

Answer 33

or if i could figure out how to make coal a clean fuel source

Answer 34

the chapter on lagrange multipliers starts: in my applications we must find the extrema of a function of several variables when the variables are restricted in some manner.

Answer 35

gimme a quick second

Answer 36

f (xyz) =xyz + (2xz + 4yz + 8xy − C) does this formula seem familiar

Answer 37

why?

Answer 38

you're end up with 2 numbers at the end, the maximum and minimum.

Answer 39

correct this is what i need

Answer 40

i need to solve for x and y

Answer 41

correct? i could be wrong

Answer 42

actually, it's a little different

Answer 43

z is negilble because it is the depth and not going to be a min or a maxium

Answer 44

the minimum and maximum values will be points on the graph which happens to be a 3 variable function

Answer 45

all the variables are going to count

Answer 46

okay

Answer 47

the maximum and minimum values will be points in a 3 dimensional space, so they'll look like (x1,y1,z1) and (x2,y2,z2)

Answer 48

if we didn't have the constraint, it would be straight forward. we'd find the partials, set them all equal to zero and then solve for the three variables.

Answer 49

since we would have fx=0, fy=0, fz=0; all equations in their own right, and we have 3 variables, we would be able to solve this problem completely.

Answer 50

depending on how many CRITICAL points we find, some could be maximum(s) and others minimum(s)

Answer 51

thoughts?

Answer 52

so take the partials and set them equal to zero and solve for a x, y, or z and plug the values into the equations

Answer 53

is that the first step?

Answer 54

that would only be if we didn't have a constraint; since we have a constraint, we need to find a way to include it in our optimization.

Answer 55

so take the constraint and solve for one of the variables. pick which ever one you'd like

Answer 56

by the way, i take it you're in college...what year?

Answer 57

i am a junior

Answer 58

where?

Answer 59

so the contratint is x^2+y^2+2z^2=25

Answer 60

and let me know once you have solved for one of the variables in the constraint...

Answer 61

at Ohio State

Answer 62

that's correct. chose one variable, x, y or z and solve for it

Answer 63

good stuff

Answer 64

I graduated from Stanford and I'm currently working in the DC area

Answer 65

chose a variable that's gonna make it easy on you cuz we'll have to plug whatever that variable is equal to into the f(x,y,z) function!

Answer 66

once we do that, we'll have one function with only 2 variables and we can find the two partials, set them equal to zero and solve!

Answer 67

show me your work and let me know what you're thinking so that I can guide you in case you get lost...

Answer 68

take your time

Answer 69

with the 2 present in the constraint the value i get solving for x=5-y-\sqrt{2z ^{2}}\]

Answer 70

\[x=5-y-\sqrt{2z ^{2}}\]

Answer 71

thank you! much better

Answer 72

let me check...

Answer 73

walk me thru how you got this step by step...or just to the part before you took the square root

Answer 74

the constraint equation is \[x ^{2}+y ^{2}+2z ^{2}=25\]

Answer 75

solve for X, so I squared both sides

Answer 76

first you took y^2 and 2z^2 to the other side

Answer 77

yeah, then squared both sides

Answer 78

and then you had \[x^2 = 25 - 2z^2 - y^2\]

Answer 79

yep

Answer 80

the square of the right isn't exactly what you got bradley

Answer 81

so should i not square the equation?

Answer 82

no, that was right

Answer 83

okay so what do i do with it? do i plug it into the partial?

Answer 84

the square root of 25-2z^2-y^2 doesn't simplify to that.

Answer 85

dude, give 10 minutes

Answer 86

okay you are fine

Answer 87

oh how i hate contraints

Answer 88

yes it does simplft to that

Answer 89

i hate this problem

Answer 90

i'm back and we're gonna solve this problem. sorry for the delay

Answer 91

i need to find a gradient of the the f(x,y,z) right?

Answer 92

yes, but we need to take the constraint into account

Answer 93

so we solve for one variable, x like you did above

Answer 94

the \[x= \sqrt{25-2z^2-y^2} \neq 5 - y - \sqrt{2x2}\]

Answer 95

the \[x=\sqrt{25-2z^2-y^2} \neq 5 - y - \sqrt{2x^2}\]

Answer 96

are we good here?

Answer 97

yes

Answer 98

if you're not sure why those two are NOT equal, look it up when you get a chance.