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

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.

Let me know when you have done that much!

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

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

yes, partials. gimme a second

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

do you have two functions?

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

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

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

brb

well, in 30

okay

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.

thanks

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

made any progress?

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

something to do with taking the gradient of the constraint

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

which calculus are you taking?

calc 4, 3-d vectors, multivarible

nice

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

i agree

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

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

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

don't worry about it

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

lol, then heading for the petroleum industry?

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

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

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.

gimme a quick second

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

why?

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

correct this is what i need

i need to solve for x and y

correct? i could be wrong

actually, it's a little different

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

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

all the variables are going to count

okay

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)

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.

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.

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

thoughts?

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

is that the first step?

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.

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

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

i am a junior

where?

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

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

at Ohio State

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

good stuff

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

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!

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!

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

take your time

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

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

thank you! much better

let me check...

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

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

solve for X, so I squared both sides

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

yeah, then squared both sides

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

yep

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

so should i not square the equation?

no, that was right

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

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

dude, give 10 minutes

okay you are fine

oh how i hate contraints

yes it does simplft to that

i hate this problem

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

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

yes, but we need to take the constraint into account

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

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

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

are we good here?

yes

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

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