anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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.
anonymous
  • anonymous
Let me know when you have done that much!
anonymous
  • anonymous
so I need to take partials in respect to each variable fx, fy, fz?

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anonymous
  • anonymous
so fx=y^2z^3 fy=2xyz^3 fz=3xy^2z^2
anonymous
  • anonymous
yes, partials. gimme a second
anonymous
  • anonymous
wait a second, rephrase your question. what do you mean by f(x,y,z)+xy^2z^3 ??
anonymous
  • anonymous
do you have two functions?
anonymous
  • anonymous
f(x,y,z)=xy^2z^3 subject to the constraint of x^2+y^2+2z^2=25
anonymous
  • 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?
anonymous
  • anonymous
not rushed it is due in the morning, thank you for your help. this site is fun
anonymous
  • anonymous
brb
anonymous
  • anonymous
well, in 30
anonymous
  • anonymous
okay
anonymous
  • 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.
anonymous
  • anonymous
thanks
anonymous
  • 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
anonymous
  • anonymous
made any progress?
anonymous
  • anonymous
i am confused because i left my book at home and I am at school
anonymous
  • anonymous
something to do with taking the gradient of the constraint
anonymous
  • anonymous
we have not done this type of problem, and i have no information in my notes
anonymous
  • anonymous
which calculus are you taking?
anonymous
  • anonymous
calc 4, 3-d vectors, multivarible
anonymous
  • anonymous
nice
anonymous
  • anonymous
problem is only worth 5 points but i think it is important to know how to do it
anonymous
  • anonymous
i agree
anonymous
  • anonymous
i'm dining and browsing pages at the same time...
anonymous
  • anonymous
in a weird way, i miss school; funny how that works
anonymous
  • anonymous
eat, i dont want to ruin your dinner, just get back to me when you are finished
anonymous
  • anonymous
don't worry about it
anonymous
  • anonymous
i only have 2 more math classes and 3 physical chem classes and 3 chem engineering classes and then I am done
anonymous
  • anonymous
lol, then heading for the petroleum industry?
anonymous
  • anonymous
i want to do desalinization, water filtration, water is the next oil!
anonymous
  • anonymous
or if i could figure out how to make coal a clean fuel source
anonymous
  • 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.
anonymous
  • anonymous
gimme a quick second
anonymous
  • anonymous
f (xyz) =xyz +  (2xz + 4yz + 8xy − C) does this formula seem familiar
anonymous
  • anonymous
why?
anonymous
  • anonymous
you're end up with 2 numbers at the end, the maximum and minimum.
anonymous
  • anonymous
correct this is what i need
anonymous
  • anonymous
i need to solve for x and y
anonymous
  • anonymous
correct? i could be wrong
anonymous
  • anonymous
actually, it's a little different
anonymous
  • anonymous
z is negilble because it is the depth and not going to be a min or a maxium
anonymous
  • anonymous
the minimum and maximum values will be points on the graph which happens to be a 3 variable function
anonymous
  • anonymous
all the variables are going to count
anonymous
  • anonymous
okay
anonymous
  • 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)
anonymous
  • 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.
anonymous
  • 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.
anonymous
  • anonymous
depending on how many CRITICAL points we find, some could be maximum(s) and others minimum(s)
anonymous
  • anonymous
thoughts?
anonymous
  • 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
anonymous
  • anonymous
is that the first step?
anonymous
  • 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.
anonymous
  • anonymous
so take the constraint and solve for one of the variables. pick which ever one you'd like
anonymous
  • anonymous
by the way, i take it you're in college...what year?
anonymous
  • anonymous
i am a junior
anonymous
  • anonymous
where?
anonymous
  • anonymous
so the contratint is x^2+y^2+2z^2=25
anonymous
  • anonymous
and let me know once you have solved for one of the variables in the constraint...
anonymous
  • anonymous
at Ohio State
anonymous
  • anonymous
that's correct. chose one variable, x, y or z and solve for it
anonymous
  • anonymous
good stuff
anonymous
  • anonymous
I graduated from Stanford and I'm currently working in the DC area
anonymous
  • 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!
anonymous
  • 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!
anonymous
  • anonymous
show me your work and let me know what you're thinking so that I can guide you in case you get lost...
anonymous
  • anonymous
take your time
anonymous
  • anonymous
with the 2 present in the constraint the value i get solving for x=5-y-\sqrt{2z ^{2}}\]
anonymous
  • anonymous
\[x=5-y-\sqrt{2z ^{2}}\]
anonymous
  • anonymous
thank you! much better
anonymous
  • anonymous
let me check...
anonymous
  • anonymous
walk me thru how you got this step by step...or just to the part before you took the square root
anonymous
  • anonymous
the constraint equation is \[x ^{2}+y ^{2}+2z ^{2}=25\]
anonymous
  • anonymous
solve for X, so I squared both sides
anonymous
  • anonymous
first you took y^2 and 2z^2 to the other side
anonymous
  • anonymous
yeah, then squared both sides
anonymous
  • anonymous
and then you had \[x^2 = 25 - 2z^2 - y^2\]
anonymous
  • anonymous
yep
anonymous
  • anonymous
the square of the right isn't exactly what you got bradley
anonymous
  • anonymous
so should i not square the equation?
anonymous
  • anonymous
no, that was right
anonymous
  • anonymous
okay so what do i do with it? do i plug it into the partial?
anonymous
  • anonymous
the square root of 25-2z^2-y^2 doesn't simplify to that.
anonymous
  • anonymous
dude, give 10 minutes
anonymous
  • anonymous
okay you are fine
anonymous
  • anonymous
oh how i hate contraints
anonymous
  • anonymous
yes it does simplft to that
anonymous
  • anonymous
i hate this problem
anonymous
  • anonymous
i'm back and we're gonna solve this problem. sorry for the delay
anonymous
  • anonymous
i need to find a gradient of the the f(x,y,z) right?
anonymous
  • anonymous
yes, but we need to take the constraint into account
anonymous
  • anonymous
so we solve for one variable, x like you did above
anonymous
  • anonymous
the \[x= \sqrt{25-2z^2-y^2} \neq 5 - y - \sqrt{2x2}\]
anonymous
  • anonymous
the \[x=\sqrt{25-2z^2-y^2} \neq 5 - y - \sqrt{2x^2}\]
anonymous
  • anonymous
are we good here?
anonymous
  • anonymous
yes
anonymous
  • anonymous
if you're not sure why those two are NOT equal, look it up when you get a chance.
anonymous
  • anonymous
i figured that part out
anonymous
  • anonymous
now back to our function, plug this new found value of x into our original function replacing it at every place where x shows up
anonymous
  • anonymous
okay i already did that
anonymous
  • anonymous
so the original function was \[f(x,y,z) = xy^2z^3\]
anonymous
  • anonymous
what did you get?
anonymous
  • anonymous
here's another thing to think about: if you had chosen y instead of x, we would be square it and getting rid of the nasty square root. you wanna do that again?
anonymous
  • anonymous
do let me know what you're thinking and if you're start getting lost...
anonymous
  • anonymous
\[\sqrt{25-2z ^{2}-y ^{2}}y ^{2}z ^{3}\]
anonymous
  • anonymous
oh did not think of that
anonymous
  • anonymous
but you see why that would have been smart?
anonymous
  • anonymous
still is...
anonymous
  • anonymous
why does the y solve and the radical go away?
anonymous
  • anonymous
oh never mind i see it, let me check
anonymous
  • anonymous
Bradley, we're solving this without using Lagrange Multipliers (LMs)....this is gonna give us another but apparently the LMs provides a nicer way....we're gonna look into that after.
anonymous
  • anonymous
it's very good to see you check your work and see why things are why they are. keep it up!
anonymous
  • anonymous
once you have the function of \[f(x,z)\] you can find fx and fz, set them equal to zero and solve them. you'll have 2 equations and 2 variables. you see why it's gonna fx and fz? because why will have been replaced by those two.
anonymous
  • anonymous
\[x(25-2z ^{2}-x ^{2})z ^{3}\]
anonymous
  • anonymous
what is this? always remember to represent functions in terms of f(x)=something...
anonymous
  • anonymous
in this case it would be \[f(x)=....\]. multiply it out and then find fx and then fz.
anonymous
  • anonymous
so what i wrote before is the f(x,z)
anonymous
  • anonymous
that's correct
anonymous
  • anonymous
now can you find fx and fz, set them equal to zero and solve for x and z?
anonymous
  • anonymous
so take the partial of each?
anonymous
  • anonymous
find each partial
anonymous
  • anonymous
note: this will give you the answer, but not using the method the suggested. i'm looking at exactly how Lagrange Multipliers work. it's something like \[\Delta f(x,y,z) = \lambda \Delta g(x,y,z)\] where \[g(x,y,z)\] will be your constraint function.
anonymous
  • anonymous
partial of fx=\[25z ^{3}-2z ^{5}x-x ^{3}z ^{3}\]
anonymous
  • anonymous
the Lagrange multiplier is no jok but we might be able to figure out if you stick around :)
anonymous
  • anonymous
fz=\[75xz ^{2}-10z ^{4}x-3x ^{3}z ^{2}\]
anonymous
  • anonymous
i am here until it is done
anonymous
  • anonymous
damn, these are some nasty looking numbers!
anonymous
  • anonymous
nasty looking functions i should say
anonymous
  • anonymous
yep, i think this is why the lagrange multiplers are needed
anonymous
  • anonymous
absolutely right
anonymous
  • anonymous
any way, let's finish this first. if we can, lol
anonymous
  • anonymous
so we set those partials to zero and see what happens
anonymous
  • anonymous
are you good with algebra?
anonymous
  • anonymous
pretty good
anonymous
  • anonymous
that's what I like to hear! so let's simply the hell out of these two and solve the system of 2 equations to find the solution
anonymous
  • anonymous
so set them both equal to zero, simplify, and then solve for x and z
anonymous
  • anonymous
let me know if you get stuck
anonymous
  • anonymous
i see alot of quadratic forumla in my future
anonymous
  • anonymous
lol
anonymous
  • anonymous
we might break half way and make the jump to langrage...
anonymous
  • anonymous
there are multiple variables so i can do the quadatic, even after factoring out some things
anonymous
  • anonymous
do you think it would be better to take the graident of f(x,z)
anonymous
  • anonymous
might be a little easier but not much easier. i'm looking at a \[5x^3-3x^2-50\] for this one
anonymous
  • anonymous
a \[5x^3-3x^2-50=0\]
anonymous
  • anonymous
a calculator might solve it...
anonymous
  • anonymous
where does that come from?
anonymous
  • anonymous
\[f _{x}=25z^3-2z^5x-x^3z^3=0 => f_{x}=25-2z^2x-x^3\]
anonymous
  • anonymous
yeah i see that
anonymous
  • anonymous
\[f_{z}=75xz^2-10z^4x-3x^3z^2=0 => f_{z}=75-10z^2-3x^2\]
anonymous
  • anonymous
the last part was \[-3x^2\]
anonymous
  • anonymous
if you solve that, eliminate \[z\] by multiplying the first with -5 and adding both equations you get what i got.
anonymous
  • anonymous
check my work, i may have made an algebraic error while I double check this langrage business
anonymous
  • anonymous
at the very least, you'll have an answer for this question and then we can worry about the method.
anonymous
  • anonymous
i am not seeing the x^3 is coming from
anonymous
  • anonymous
in which partial?
anonymous
  • anonymous
x or z?
anonymous
  • anonymous
no when i added the 2 i got 12x^2 - 50
anonymous
  • anonymous
in the fx why did you not factor out z^3
anonymous
  • anonymous
I did. look at the part after the => sign
anonymous
  • anonymous
am I missing something here? plz do let me know, I could have easily made a mistake...
anonymous
  • anonymous
btw, I admire your persistence. I know you could say the same for me but I enjoy this kind of thing. Usually I enjoy working on challenging and interesting problems, but I didn't when I was a student. I did, but I didn't put in the work. so good job thus far.
anonymous
  • anonymous
by the way, I have a problem here in my book which is a superb example for solving this problem with lagrange multipliers. I don't have a scan, but I can take a pic with my camera. so if you'd like you can send your email so i can forward it to you.
anonymous
  • anonymous
lets start with the basics f(x,z)=\[25xz^3-2z^5x-x^3z^3\]
anonymous
  • anonymous
anonymous
  • anonymous
fx=\[25z^3-2z^5-3x^2z^3\]
anonymous
  • anonymous
fx=\[z^3(25-2z^2-3x^2)\]
anonymous
  • anonymous
btw, what you typed in earlier: f (xyz) =xyz + (2xz + 4yz + 8xy − C) does this formula seem familiar. this is what we'll end up referring back to for the langrage.
anonymous
  • anonymous
was that in your notes somewhere?
anonymous
  • anonymous
fz=\[75xz^2-10z^4x-3x^3z^2\]
anonymous
  • anonymous
that was from a volume equation the variables were off, dont think it matters
anonymous
  • anonymous
fz=\[xz^2(75-10z^2-3x^2)\]
anonymous
  • anonymous
i see the problem in the book but i guess i am having trouble seeing what to do with what i find
anonymous
  • anonymous
so you have the example to a similar problem as well that is being solved using Lagrange Multipliers?
anonymous
  • anonymous
by the way, the x^3 came from the original fx that you gave me. I didn't check it. if it was true, that's where the x^3 was coming from.
anonymous
  • anonymous
must have been a type, there shouldn't be a x^3. just checked.
anonymous
  • anonymous
yea, so it should nicely reduce then
anonymous
  • anonymous
okay
anonymous
  • anonymous
so i got 12x^2-50
anonymous
  • anonymous
so then it is x=\[\sqrt{50}/12\]
anonymous
  • anonymous
\[5\sqrt{2}/12\]
anonymous
  • anonymous
what about z?
anonymous
  • anonymous
by adding-5 to fx and adding to fz the z^2 canceled
anonymous
  • anonymous
right, but now you plug the value of x into one of them to solve for z as well. remember?
anonymous
  • anonymous
sorrying multiply -5 to fx
anonymous
  • anonymous
so plug in the \[5\sqrt{2}/12\] to the orignal functions
anonymous
  • anonymous
and solve for z?
anonymous
  • anonymous
right
anonymous
  • anonymous
this number is nasty
anonymous
  • anonymous
it shouldn't be
anonymous
  • anonymous
maybe i am doing poor algebra, wait do i plug it into f(x,y,z) or f(x,z)
anonymous
  • anonymous
into f(x,z)
anonymous
  • anonymous
\[f_{x}=25-2z^2-3x^2\] and \[f_{z}=75-10z^2-3x^2\]
anonymous
  • anonymous
\[\sqrt{50}/12(25-2z^2-(\sqrt{50}/12)^2)z^3\]
anonymous
  • anonymous
subtract those two: \[f_{z}-f_{x} = 50-8z^2\]
anonymous
  • anonymous
haha so much easier
anonymous
  • anonymous
5/2
anonymous
  • anonymous
that's right. now plug that into either fx or fz
anonymous
  • anonymous
so if i plugged the z value into fx, i would solve for X?
anonymous
  • anonymous
right
anonymous
  • anonymous
what do you have for me?
anonymous
  • anonymous
you realize that once you get the x we'll now have x and z, and remember the expression of y you got earlier by solving for y? well, if you go and plug this x and z in it, you'll have the y which will give you a point in 3D (x,y,z) and that will be the solution!
anonymous
  • anonymous
I just got through looking at the Lagrange and it is considerably easier once you get it. let me know when you're ready to move on.
anonymous
  • anonymous
the x is difficult because of the fractions
anonymous
  • anonymous
It's nearing 11 o'clock my time, so we'll have to hurry up a bit :)
anonymous
  • anonymous
awww, not a fan of fractions are ya? no worries
anonymous
  • anonymous
just teasin' by the way,
anonymous
  • anonymous
i agree i am about done
anonymous
  • anonymous
so let's figure out this fraction business
anonymous
  • anonymous
\[f_{x,z=5/2}=25-2(5/2)^2-3x^2x\]
anonymous
  • anonymous
simplying gives us: \[0=25-25/2-3x^2=50/2-25/2-3x^3\]
anonymous
  • anonymous
50/2-25/2=25/2, right?
anonymous
  • anonymous
because you get (50-25)/2=25/2
anonymous
  • anonymous
so we now have \[25/2-3x^2=0 => 25/2=3x^2\]
anonymous
  • anonymous
you can get x from there
anonymous
  • anonymous
\[\{5}/6\]
anonymous
  • anonymous
5/6
anonymous
  • anonymous
not exactly...
anonymous
  • anonymous
sorry my brain is fried the whole thing must be squared
anonymous
  • anonymous
exactly
anonymous
  • anonymous
don't worry, that makes two of us
anonymous
  • anonymous
\[\sqrt{25/6}\]
anonymous
  • anonymous
which simplifies to...
anonymous
  • anonymous
\[5/\sqrt{6}\]
anonymous
  • anonymous
yes, but by convention you don't leave a radical in the bottom so you...
anonymous
  • anonymous
multiply both top and both by the same radical and you get...
anonymous
  • anonymous
\[5\sqrt{6}/6\]
anonymous
  • anonymous
that's it
anonymous
  • anonymous
now, if we didn't make any stupid mistakes along the way, those two should be two of the 3 you need to make a point in 3D
anonymous
  • anonymous
now plug all three into the equation
anonymous
  • anonymous
now plug those into the y expression you got earlier and you should get the third point to form (x,y,z)
anonymous
  • anonymous
once you have the point, that's it. that should be the optimal point you were looking for to satisfy whatever the problem was asking you to satisfy in real life
anonymous
  • anonymous
now onto lagrange: given \[f(x,y,z)\], and the constraint \[x^2+y^2+2z^2=25\], let \[g(x,y,z)=x^2+y^2+2z^2-25\]. Now, all the wise Lagrange himself said was that given that the gradient of \[g(x,y,z\] neve equals zero, there exist a constant such that the gradient of \[f(x,y,z)\] equals the constant times the gradient of \[g(x,y,z)\].
anonymous
  • anonymous
okay
anonymous
  • anonymous
first, you know now how you're gonna get the y?
anonymous
  • anonymous
and at least have your optimal point? before we do this now. we're just gonna quickly set up the equations and I'm gonna hit the pellets
anonymous
  • anonymous
*sheets
anonymous
  • anonymous
yeah y=\[\sqrt{25/3}\]
anonymous
  • anonymous
or \[5\sqrt{3}/3\]
anonymous
  • anonymous
sweet. i haven't checked it. make sure all the algebra's right.
anonymous
  • anonymous
go ahead and go to sleep i appreciate all your help, it was above the call of duty
anonymous
  • anonymous
do you see how I wrote the constraint as the function g(x,y,z) ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so langrage gives you a bunch of easy equations to solve: here they are:
anonymous
  • anonymous
he says that there exists a constant, \[lambda\] that when multiplied by the constraint, you get the solution. interesting man
anonymous
  • anonymous
okay
anonymous
  • anonymous
so here are the equations: fx = lambda*gx, fy = lambda*gy, fz = lambda*gz, and finally, we have our constraint: g(x)=0;
anonymous
  • anonymous
lambda as I am using it here is of course \[\lambda\]
anonymous
  • anonymous
with the g(x,y,z)=0 we have 4 equations which is great because we have 4 variables: x,y,z, and \[\lambda\]
anonymous
  • anonymous
so take the partial of gx, gy, gz and solve for lambda
anonymous
  • anonymous
that is exactly correct! of course you'll also be taking partials of fx,fy,and fz but remember that here we're talking about the original \[f(x,y,z)=xy^2z3\]
anonymous
  • anonymous
yes! awesome, i got it! thank you so much for your help
anonymous
  • anonymous
in the process you'll solve for the lambda, x,y, and z. you're interested in the x,y,and z and if we did everything right, they should match what we got earlier.
anonymous
  • anonymous
i understand it know thanks so much
anonymous
  • anonymous
good to know where everything comes from
anonymous
  • anonymous
or that there are easier ways of getting things done :)
anonymous
  • anonymous
if i ever get stuck can i email you?
anonymous
  • anonymous
you're welcome. you have my email, shoot me one if you ever need some more help. i don't visit here often.
anonymous
  • anonymous
you beat me to it; absolutely.
anonymous
  • anonymous
thank you very much
anonymous
  • anonymous
aite man, ciao
anonymous
  • anonymous
sleep well
anonymous
  • anonymous
thank you

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