October 16th, 2018, 09:22 PM  #1 
Newbie Joined: Aug 2018 From: Suisun City, CA Posts: 7 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? 
October 16th, 2018, 10:04 PM  #2 
Newbie Joined: Aug 2018 From: Suisun City, CA Posts: 7 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 11:21 AM. 
October 17th, 2018, 08:33 AM  #3 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
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. 
October 17th, 2018, 10:02 AM  #4 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  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.

October 17th, 2018, 10:37 AM  #5 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
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. Last edited by skipjack; October 17th, 2018 at 11:20 AM. 
October 17th, 2018, 03:41 PM  #6 
Newbie Joined: Aug 2018 From: Suisun City, CA Posts: 7 Thanks: 0  
October 18th, 2018, 01:28 PM  #7 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  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. 

Tags 
lagrange, multipliers 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Lagrange Multipliers  Ksci  Calculus  1  September 14th, 2015 05:16 PM 
Lagrange Multipliers  yo79  Math Events  2  March 30th, 2013 01:00 AM 
Lagrange multipliers  aaronmath  Calculus  1  March 23rd, 2012 10:53 AM 
need help in Lagrange multipliers !!!!!  Tiome_nguyen  Calculus  0  November 1st, 2011 10:54 PM 
Lagrange Multipliers  ziggy41  Calculus  5  October 22nd, 2010 09:31 PM 