My Math Forum Lagrange multiplier (am I doing this right?)?

 Applied Math Applied Math Forum

 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+6x-2xy-?(x-y²) L'x = 6-2y-?=0 L'y = 18-2x-2?y=0 constraint: x-y²=0 This means: x=y² y= ?x or y=-?x 18-2x-2?y = 0 18-2y²-2?y=0 ?=(18-2y²)/2y = (6-2y)/1 18-2y² = 12y-4y² 18= 12y-2y² This gives me: -2y²+12y-18=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 $z=f(x,y)$ subject to the constraint $g(x,y)=0$, solve the system of equations: $f_x(x,y)=\lambda g_x(x,y)$ $f_y(x,y)=\lambda g_y(x,y)$ $g(x,y)=0$ In our case, we have: $f(x,y)=18y+6x-2xy$ and $g(x,y)=x-y^2$ so we solve the system: $6-2y=\lambda$ $18-2x=-2\lambda y$ $x-y^2=0$ From the first 2 equations, we see: $\lambda=6-2y=\frac{x-9}{y}$ $x=6y-2y^2+9$ Substituting for x into the 3rd equation, we get: $6y-2y^2+9-y^2=0$ $6y-3y^2+9=0$ $y^2-2y-3=0$ $(y-3)(y+1)=0$ i) When $y=-1$ we have: $x=1$ and $\lambda=8$ and $f$$1,-1$$=-10$ This is the constrained minimum. ii) When $y=3$ we have: $x=9$ and $\lambda=0$ and $f$$9,3$$=54$ 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

 Tags lagrange, multiplier

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post MMCS Calculus 3 December 7th, 2013 07:44 AM Brazen Calculus 1 January 15th, 2013 10:29 AM dk1702 Abstract Algebra 1 July 21st, 2010 05:25 AM OSearcy4 Calculus 2 October 16th, 2009 01:44 PM roonaldo17 Calculus 0 November 16th, 2008 11:27 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top