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May 2nd, 2015, 05:23 AM  #1 
Newbie Joined: May 2015 From: Imperium Romanum Posts: 13 Thanks: 0  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! 
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. 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!


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