My Math Forum Need help with Lagrange multiplier...

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May 2nd, 2015, 05:23 AM   #1
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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.)?

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 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!

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