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 Calculus Calculus Math Forum

May 2nd, 2015, 05:23 AM   #1
Newbie

Joined: May 2015
From: Imperium Romanum

Posts: 13
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Need help with Lagrange multiplier...

Hi Everyone,

How do I workout the partial derivative for 1.) and the workings to the solution 2.)?

Your help is much appreciated!
Attached Images 1111.jpg (75.6 KB, 9 views) May 2nd, 2015, 06:21 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Write $\displaystyle f(x)= \sqrt{x^2+ y^2}= (x^2+ y^2)^{1/2}$. Think of that as $\displaystyle f(u)=u^{1/2}$ and $\displaystyle u(x,y)= x^2+ y^2$. By the chain rule, $\displaystyle \frac{\partial f}{\partial x}= \frac{df}{du}\frac{\partial u}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}= \frac{df}{du}\frac{\partial u}{\partial y}$ $\displaystyle \frac{df}{du}= (1/2)u^{1/2- 1}= (1/2)u^{-1/2}$, $\displaystyle \frac{\partial u}{\partial x}= 2x$, and $\displaystyle \frac{\partial u}{\partial y}= 2y$ So $\displaystyle \frac{\partial f}{\partial x}= (1/2)u^{-1/2}(2x)$$\displaystyle = x(x^2+ y^2)^{-1/2}$$\displaystyle = \frac{x}{\sqrt{x^2+ y^2}}$ and $\displaystyle \frac{\partial f}{\partial y}= (1/2)u^{-1/2}(2y)= y(x^2+ y^2)^{-1/2}= \frac{y}{\sqrt{x^2+ y^2}}$ For your second question, since $\displaystyle 2+ \frac{1}{\sqrt{x^2+ y^2}}$ is never negative, you can divide both sides by it, leaving x= y. You have the constraint x+ y= 1 so that x= y= 1/2. Thanks from Caesar95 Last edited by Country Boy; May 2nd, 2015 at 06:31 AM. May 2nd, 2015, 08:04 AM #3 Newbie   Joined: May 2015 From: Imperium Romanum Posts: 13 Thanks: 0 Thank you so much! Tags lagrange, multiplier Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Brazen Calculus 1 January 15th, 2013 10:29 AM trey01 Applied Math 2 March 25th, 2012 07:14 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

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