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 October 1st, 2017, 06:23 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 165 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, 08:24 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,940 Thanks: 794 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 $\displaystyle \left(-\frac{3}{2}\sqrt{2}, -\frac{3}{2}\sqrt{2}\right)$. Thanks from zollen
October 1st, 2017, 02:43 PM   #3
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 Originally Posted by zollen 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.
Your condition implies that the expression under the square root is always 0. So You need a min and max for f(x,y)=xy under the condition.

October 1st, 2017, 02:45 PM   #4
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The constraint was x^2 + y^2 <= 9

The official answers are: 5 and -9/2

Quote:
 Originally Posted by mathman Your condition implies that the expression under the square root is always 0. So You need a min and max for f(x,y)=xy under the condition.

 October 1st, 2017, 03:46 PM #5 Senior Member     Joined: Sep 2015 From: USA Posts: 1,752 Thanks: 894 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! Thanks from zollen
October 1st, 2017, 04:11 PM   #6
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Does solving the general problem x=r cos(theta) also includes the boundary?

Quote:
 Originally Posted by romsek 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, 04:18 PM #7 Senior Member     Joined: Sep 2015 From: USA Posts: 1,752 Thanks: 894 you end up with a singularity in the general problem when $r=3$ which causes you to miss the minima
 October 1st, 2017, 05:10 PM #8 Senior Member   Joined: Jan 2017 From: Toronto Posts: 165 Thanks: 2 $\displaystyle f(x, y, \lambda) = xy + \sqrt{9-x^2-y^2} - \lambda ( x^2 +y^2 - 9)$ * 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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