My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree1Thanks
  • 1 Post By zylo
Reply
 
LinkBack Thread Tools Display Modes
October 16th, 2018, 10:22 PM   #1
Newbie
 
Joined: Aug 2018
From: Suisun City, CA

Posts: 6
Thanks: 0

Lagrange Multipliers

$\displaystyle f(x,y,z)=8x+4y+4z$ ; subject to $\displaystyle 4x^2+4y^2+4z^2=36$

I have found the partial derivatives of each and found the value of λ for each partial.

$\displaystyle fx: 8=λ8x$ ==> $\displaystyle λ=1/x$
$\displaystyle fy: 4=λ8y$ ==> $\displaystyle λ=1/2y$
$\displaystyle fz: 4=λ8z$ ==> $\displaystyle λ=1/2z$
$\displaystyle 4x^2+4y^2+4z^2=36$

$\displaystyle 1/x=1/2y=1/2z$

This is as far as I am able to get. I'm not having any issues with problems of two variables, but now that I am at three variables, I am stuck. Could someone give me a push in the right direction?
cvbody is offline  
 
October 16th, 2018, 11:04 PM   #2
Newbie
 
Joined: Aug 2018
From: Suisun City, CA

Posts: 6
Thanks: 0

$\displaystyle x=2y=2z$
Plugging everything into the constraint for y gave me
$\displaystyle 4(2y)^2+4y^2+2(2y)^2=36$
$\displaystyle 16y^2+4y^2+8y^2=36$
$\displaystyle 28y^2=36$
when solved..
$\displaystyle y=(\pm)(3 √7)/7$

When using these values to solve for x and z, then plugging those into $\displaystyle f(x,y,z)$the values I get for the max and min are
$\displaystyle (72 √7)/7$ and $\displaystyle -(72 √7)/7$ which are incorrect.

Last edited by skipjack; October 17th, 2018 at 12:21 PM.
cvbody is offline  
October 17th, 2018, 09:33 AM   #3
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,603
Thanks: 115

df=8dx+4dy+4dz=0, but dx,dy,dz, aren't independent. They are subject to constraint:

8xdx+8ydy+8zdz=0. Solve for dz, sub in previous equation and then set coeffients of dx and dy to zero to get:

x=-2z, y=-z and substitute that into constraint to get z=$\displaystyle \pm$ sqrt(3/2}

That's the principle.
zylo is offline  
October 17th, 2018, 11:02 AM   #4
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,603
Thanks: 115

Quote:
Originally Posted by cvbody View Post
$\displaystyle x=2y=2z$
When using these values to solve for x and z, then plugging those into $\displaystyle f(x,y,z)$the values I get for the max and min are
$\displaystyle (72 √7)/7$ and $\displaystyle -(72 √7)/7$ which are incorrect.
You did everything right. LM just gives you locations of extrema. There are rules for determining whether you have a max, min, or saddle. I don't remember and don't feel like looking them up. You can do it too.
zylo is offline  
October 17th, 2018, 11:37 AM   #5
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,603
Thanks: 115

You might check out:


Just substitute your 2 vals into f to get max and min. I don't think test for saddle points relevant here. Don't go by my algebra, I usually get it wrong.
Thanks from cvbody

Last edited by skipjack; October 17th, 2018 at 12:20 PM.
zylo is offline  
October 17th, 2018, 04:41 PM   #6
Newbie
 
Joined: Aug 2018
From: Suisun City, CA

Posts: 6
Thanks: 0

Quote:
Originally Posted by zylo View Post
You might check out:


Just substitute your 2 vals into f to get max and min. I don't think test for saddle points relevant here. Don't go by my algebra, I usually get it wrong.
I forgot all about this guy! Thanks for the reminder!
cvbody is offline  
October 18th, 2018, 02:28 PM   #7
Senior Member
 
Joined: Mar 2015
From: New Jersey

Posts: 1,603
Thanks: 115

Lagrange

The question of saddle points in the case of Lagrange Multipliers bothered me since I couldn't find anything on it. So


Assume you find critial points of f(x,y,z) subject to (constrained by) g(x;y,z)=0. Are they local max/min or saddlepoints?

$\displaystyle g_{x}dx+g_{y}dy+g_{z}dz=0 \rightarrow$
$\displaystyle dz=-(g_{x}/g_{z})dx-(g_{y}/g_{z})dy $
$\displaystyle F(x,y) = f(x,y,z(x,y))$
$\displaystyle dF=f_{x}dx+f_{y}dy+f_{z}dz$
$\displaystyle dF=(f_{x}-g_{x}\frac{f_{z}}{g_{z}})dx + (f_{y}-g_{y}\frac{f_{z}}{g_{z}})dy = F_{x}dx+F_{y}dy$
$\displaystyle F_{x} = f_{x}-g_{x}\frac{f_{z}}{g_{z}}=h(x,y,z)$
$\displaystyle dF_{x} = (h_{x}-g_{x}\frac{h_{z}}{g_{z}})dx + (h_{y}-g_{y}\frac{h_{z}}{g_{z}})dy$
$\displaystyle F_{xx}=(h_{x}-g_{x}\frac{h_{z}}{g_{z}})$
$\displaystyle F_{xy}= (h_{y}-g_{y}\frac{h_{z}}{g_{z}})$

$\displaystyle F_{y} = f_{y}-g_{y}\frac{f_{z}}{g_{z}}=r(x,y,z)$
$\displaystyle dF_{y} = (r_{x}-g_{x}\frac{r_{z}}{g_{z}})dx + (r_{y}-g_{y}\frac{r_{z}}{g_{z}})dy$
$\displaystyle F_{yy}= (r_{y}-g_{y}\frac{r_{z}}{g_{z}}$)

You are left with a mess of straight forward partial derivatives which you can evaluate at critical points to determine $\displaystyle F_{xx}, F_{yy}, F_{xy}$ at critical points. Assumed $\displaystyle F_{xy}=F_{yx}$ but didn't check.

See https://en.wikipedia.org/wiki/Second...erivative_test

In principle, you could solve g for z, sub in f, and then find extrema and test for saddle points of a function of two variables. Partial derivatives may be easier to find than solving g(x,y,z) for z, the reason for Lagrange multipliers.
zylo is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
lagrange, multipliers



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Lagrange Multipliers Ksci Calculus 1 September 14th, 2015 06:16 PM
Lagrange Multipliers yo79 Math Events 2 March 30th, 2013 02:00 AM
Lagrange multipliers aaron-math Calculus 1 March 23rd, 2012 11:53 AM
need help in Lagrange multipliers !!!!! Tiome_nguyen Calculus 0 November 1st, 2011 11:54 PM
Lagrange Multipliers ziggy41 Calculus 5 October 22nd, 2010 10:31 PM





Copyright © 2018 My Math Forum. All rights reserved.