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October 1st, 2017, 06:23 AM   #1
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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.
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October 1st, 2017, 08:24 AM   #2
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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 $\displaystyle \left(-\frac{3}{2}\sqrt{2}, -\frac{3}{2}\sqrt{2}\right)$.
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October 1st, 2017, 02:43 PM   #3
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Quote:
Originally Posted by zollen View Post
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.
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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 View Post
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.
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October 1st, 2017, 03:46 PM   #5
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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!
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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 View Post
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!
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October 1st, 2017, 04:18 PM   #7
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you end up with a singularity in the general problem when $r=3$ which causes you to miss the minima
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October 1st, 2017, 05:10 PM   #8
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$\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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