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? 
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. 
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. 
October 17th, 2018, 11:02 AM  #4 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,603 Thanks: 115  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, 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. Last edited by skipjack; October 17th, 2018 at 12:20 PM. 
October 17th, 2018, 04:41 PM  #6 
Newbie Joined: Aug 2018 From: Suisun City, CA Posts: 6 Thanks: 0  
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. 

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