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March 19th, 2012, 11:10 AM  #1 
Newbie Joined: Mar 2012 Posts: 6 Thanks: 0  Lagrange multiplier (am I doing this right?)?
Hi guy's, could someone please tell me if I'm doing this right!? the question that was asked: f (x, y) = 18y + 6x ? 2xy Calculate (by Lagrange multiplier) the stationary points of f(x,y), under the constraint: x ? y² = 0 Also calculate the value of the function at the stationary points and also calculate 'Lagrange multiplier' (?) for the(se) points. So what i did: L(x,y,?)= 18y+6x2xy?(xy²) L'x = 62y?=0 L'y = 182x2?y=0 constraint: xy²=0 This means: x=y² y= ?x or y=?x 182x2?y = 0 182y²2?y=0 ?=(182y²)/2y = (62y)/1 182y² = 12y4y² 18= 12y2y² This gives me: 2y²+12y18=0 On this is used the ABC formula which gave me: X(1)= 3 and X(2)= 3. Meaning x = 3 So y=?3 or y=?3 Corresponding points (3,?3) (3,?3) I would really really appreciate it if someone could check this for me!! Thank you for your attention p.s. sorry if its a bit messy, and my apologies if i have posted this in the wrong section. 
March 19th, 2012, 12:05 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs  Re: Lagrange multiplier (am I doing this right?)?
To find the extrema of subject to the constraint , solve the system of equations: In our case, we have: and so we solve the system: From the first 2 equations, we see: Substituting for x into the 3rd equation, we get: i) When we have: and and This is the constrained minimum. ii) When we have: and and This is the constrained maximum. 
March 25th, 2012, 07:14 AM  #3 
Newbie Joined: Mar 2012 Posts: 6 Thanks: 0  Re: Lagrange multiplier (am I doing this right?)?
Thank you soooooo much Mark!! Can´t thank you enough 

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