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October 1st, 2017, 05:23 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 140 Thanks: 2  Another Lagrange Multipliers problem
Find the maximum and minimum values of $\displaystyle f(x, y) = xy +\sqrt { 9  x^2  y^2 } $ when $\displaystyle x^2 + y^2 = 9.$ The question seems straight forward but I am having problem getting the values of x and y. I would be much appreciated if you could show me how you get the x and y values. 
October 1st, 2017, 07:24 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,719 Thanks: 699 
Well, lets see: $\displaystyle \nabla f= \nabla xy+ \sqrt{9 x^2 y^2}= (y+ (1/2)(9 x^2 y^2)^{1/2}(2x)\vec{i}+ (x+ (1/2)(9 x^2 y^2)^{1/2}(2y)\vec{j}= (y x/\sqrt{9 x^2 y^2})\vec{i}+ (x y/\sqrt{9 x^2 y^2})\vec{j}$ $\displaystyle \nabla x^2+ y^2= 2x\vec{i}+ 2y\vec{j}$. At a max or min, there must be a constant, $\displaystyle \mu$, the "Lagrange multiplier" such that $\displaystyle (y x/\sqrt{9 x^2 y^2})\vec{i}+ (x y/\sqrt{9 x^2 y^2})\vec{j}= \lambda (2x\vec{i}+ 2y\vec{j})$. So we must have $\displaystyle y x/\sqrt{9 x^2 y^2}= 2\lambda x$ and $\displaystyle x y/\sqrt{9 x^2 y^2}= 2\lambda y$. Since a specific value for $\displaystyle \lambda$ is not necessary for a solution to the problem, I find it often best to first eliminate $\displaystyle \lambda$ by dividing one equation by the other. Here that gives $\displaystyle \frac{y \frac{x}{\sqrt{9 x^2 y^2}}}{x \frac{y}{\sqrt{9 x^2 y^2}}}= \frac{y}{x}$ So $\displaystyle xy \frac{x^2}{\sqrt{9 x^2 y^2}}= xy \frac{y^2}{\sqrt{9 x^2 y^2}}$ $\displaystyle x^2= y^2$ so that either $\displaystyle y= x$ or $\displaystyle y= x$. Putting either $\displaystyle y= x$ or $\displaystyle y= x$ into the condition that $\displaystyle x^2+ y^2= 9$ gives $\displaystyle 2x^2= 9$. $\displaystyle x= \pm\frac{3}{\sqrt{2}}= \pm\frac{3\sqrt{2}}{2}$. There are four possible answers: $\displaystyle \left(\frac{3}{2}\sqrt{2}, \frac{3}{2}\sqrt{2}\right)$, $\displaystyle \left(\frac{3}{2}\sqrt{2}, \frac{3}{2}\sqrt{2}\right)$, $\displaystyle \left(\frac{3}{2}\sqrt{2}, \frac{3}{2}\sqrt{2}\right)$, and $\displaystyle \left(\frac{3}{2}\sqrt{2}, \frac{3}{2}\sqrt{2}\right)$. 
October 1st, 2017, 01:43 PM  #3  
Global Moderator Joined: May 2007 Posts: 6,341 Thanks: 532  Quote:
 
October 1st, 2017, 01:45 PM  #4 
Senior Member Joined: Jan 2017 From: Toronto Posts: 140 Thanks: 2  
October 1st, 2017, 02:46 PM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749 
what you have to do is solve the general problem for $r \in [0,3)$ and then also solve the problem where $r=3$ I.e. you look for extrema on the interior, and then on the boundary. When you do this you'll catch em all! 
October 1st, 2017, 03:11 PM  #6 
Senior Member Joined: Jan 2017 From: Toronto Posts: 140 Thanks: 2 
Does solving the general problem x=r cos(theta) also includes the boundary? 
October 1st, 2017, 03:18 PM  #7 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749 
you end up with a singularity in the general problem when $r=3$ which causes you to miss the minima

October 1st, 2017, 04:10 PM  #8 
Senior Member Joined: Jan 2017 From: Toronto Posts: 140 Thanks: 2 
$\displaystyle f(x, y, \lambda) = xy + \sqrt{9x^2y^2}  \lambda ( x^2 +y^2  9) $ * or * $\displaystyle f(r, \theta, \lambda) = 1/2 r^2 sin( 2 \theta ) + \sqrt{ 9  r^2 }  \lambda ( r^2  9 ) $ Both approaches yeild the same result: $\displaystyle ( 3/\sqrt{2} , 3/\sqrt{2} , 9/2 ) $ $\displaystyle ( 3/\sqrt{2} , 3/\sqrt{2} , 9/2 ) $ $\displaystyle ( 3/\sqrt{2} , 3/\sqrt{2} , 9/2 ) $ $\displaystyle ( 3/\sqrt{2} , 3/\sqrt{2} , 9/2 ) $ How do I formulate the special case for the local max? (Answer: 5) 

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