Mathematics 23 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

Let me know when you have done that much!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes, partials. gimme a second

OpenStudy (anonymous):

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

OpenStudy (anonymous):

do you have two functions?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

brb

OpenStudy (anonymous):

well, in 30

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

something to do with taking the gradient of the constraint

OpenStudy (anonymous):

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

OpenStudy (anonymous):

which calculus are you taking?

OpenStudy (anonymous):

calc 4, 3-d vectors, multivarible

OpenStudy (anonymous):

nice

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i agree

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

lol, then heading for the petroleum industry?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

gimme a quick second

OpenStudy (anonymous):

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

OpenStudy (anonymous):

why?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

correct this is what i need

OpenStudy (anonymous):

i need to solve for x and y

OpenStudy (anonymous):

correct? i could be wrong

OpenStudy (anonymous):

actually, it's a little different

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

all the variables are going to count

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

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)

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thoughts?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

is that the first step?

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i am a junior

OpenStudy (anonymous):

where?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

at Ohio State

OpenStudy (anonymous):

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

OpenStudy (anonymous):

good stuff

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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!

OpenStudy (anonymous):

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!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thank you! much better

OpenStudy (anonymous):

let me check...

OpenStudy (anonymous):

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

OpenStudy (anonymous):

the constraint equation is $x ^{2}+y ^{2}+2z ^{2}=25$

OpenStudy (anonymous):

solve for X, so I squared both sides

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yeah, then squared both sides

OpenStudy (anonymous):

and then you had $x^2 = 25 - 2z^2 - y^2$

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so should i not square the equation?

OpenStudy (anonymous):

no, that was right

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

dude, give 10 minutes

OpenStudy (anonymous):

okay you are fine

OpenStudy (anonymous):

oh how i hate contraints

OpenStudy (anonymous):

yes it does simplft to that

OpenStudy (anonymous):

i hate this problem

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes, but we need to take the constraint into account

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

are we good here?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

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