Discovery Of Max

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#1 h Oct 21, 2010, 6:41 PM • 2 Y

Y by Adventure10, Mango247

Hello.

$x$ , $y$ are real numbers, where $x^{2}+4y^{2}=4$
Find Max (Min) of $x^{2}+xy-2y^{2}$ without trigonometry.

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1008 posts

kuing

#2 h Oct 22, 2010, 10:51 AM • 2 Y

Y by Adventure10, Mango247

Headhunter wrote:

Hello.

$x$ , $y$ are real numbers, where $x^{2}+4y^{2}=4$
Find Max (Min) of $x^{2}+xy-2y^{2}$ without trigonometry.

$x^2 + 4y^2 = 4$ , let $x^2 + xy - 2y^2 = f$ , then

$\Rightarrow k\left( {x^2 + xy - 2y^2 } \right) + x^2 + 4y^2 = 4 + kf$
$\Leftrightarrow \left( {k + 1} \right)x^2 + kxy + \left( {4 - 2k} \right)y^2 = 4 + kf.$

let $4\left( {k + 1} \right)\left( {4 - 2k} \right) = k^2 \Rightarrow k = \frac{4}{9}\left( {1 \pm \sqrt {10} } \right).$

by Cauchy-Schwarz Inequality, we have

(1), $k = \frac{4}{9}\left( {1 + \sqrt {10} } \right)$

$\Rightarrow 4 + \frac{4}{9}\left( {1 + \sqrt {10} } \right)f = \frac{1}{9}\,\left( {13 + 4\sqrt {10} } \right)\left( {x + \left( {2\sqrt {10} - 6} \right)y} \right)^2$

　 $\le \frac{1}{9}\,\left( {13 + 4\sqrt {10} } \right)\left( {x^2 + 4y^2 } \right)\left( {1 + \frac{{\left( {2\sqrt {10} - 6} \right)^2 }}{4}} \right)$

$\Rightarrow f \le \frac{{\frac{1}{9}\,\left( {13 + 4\sqrt {10} } \right)\left( {4 + \left( {2\sqrt {10} - 6} \right)^2 } \right) - 4}}{{\frac{4}{9}\left( {1 + \sqrt {10} } \right)}} = 1 + \sqrt {10} \;;$

(2), $k = \frac{4}{9}\left( {1 - \sqrt {10} } \right)$

$\Rightarrow 4 + \frac{4}{9}\left( {1 - \sqrt {10} } \right)f = \frac{1}{9}\,\left( {13 - 4\sqrt {10} } \right)\left( {x - \left( {2\sqrt {10} + 6} \right)y} \right)^2$

　 $\le \frac{1}{9}\,\left( {13 - 4\sqrt {10} } \right)\left( {x^2 + 4y^2 } \right)\left( {1 + \frac{{\left( {2\sqrt {10} + 6} \right)^2 }}{4}} \right)$

$\Rightarrow f \ge \frac{{\frac{1}{9}\,\left( {13 - 4\sqrt {10} } \right)\left( {4 + \left( {2\sqrt {10} + 6} \right)^2 } \right) - 4}}{{\frac{4}{9}\left( {1 - \sqrt {10} } \right)}} = 1 - \sqrt {10} \; .$

easy to know the equalty is can holds, so max and min is $1+\sqrt{10}$ and $1-\sqrt{10}$ .

hope this solution is right.

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#3 h Oct 22, 2010, 5:45 PM • 2 Y

Y by Adventure10, Mango247

$x^2+4y^2=4$ is a simple ellipse in the xy-plane.

Let $z=x^2+xy-2y^2$ , look at this as a 3d graph,
it's just a simple rotated hyperbolic paraboloid (because of the "xy" term)

$z=(x+2y)(x-y)$
$z=0$ when $y=- \frac{1}{2}x$ and $y=x$
thus
the max will occur when $y=\frac{1}{4}x$
the min will occur when $y=-4x$
(and these values are global)
to help you see this, here are:

3d plot of $z=x^2+xy-2y^2$ :
//cdn.artofproblemsolving.com/images/4048e91ba033ec1cc6312d80c5a5b52050d7969c.gif

//cdn.artofproblemsolving.com/images/4048e91ba033ec1cc6312d80c5a5b52050d7969c.gif

Contour plot $z=x^2+xy-2y^2$ :
//cdn.artofproblemsolving.com/images/c96a545227b5598a4120f9d1b91c92b24815b4cb.gif

//cdn.artofproblemsolving.com/images/c96a545227b5598a4120f9d1b91c92b24815b4cb.gif

on the ellipse $x^2+4y^2=4$ in the xy-plane
the max will occur at the intersection of $x^2+4y^2=4$ and $y=\frac{1}{4}x \rightarrow (\frac{8}{\sqrt{17}},\frac{1}{\sqrt{17}})$ and $(-\frac{8}{\sqrt{17}},-\frac{1}{\sqrt{17}})$
the min will occur at the intersection of $x^2+4y^2=4$ and $y=-4x \rightarrow (\frac{2}{\sqrt{65}}, -\frac{8}{\sqrt{65}})$ and $(-\frac{2}{\sqrt{65}}, \frac{8}{\sqrt{65}})$

max $z=\frac{70}{17}$
min $z=-\frac{76}{65}$

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985 posts

#4 h Oct 23, 2010, 2:27 AM • 2 Y

Y by Adventure10, Mango247

Cant you just easily find the derivative of $4-4y^2+2y\sqrt{1-y^2}-2y^2$ and set it equal to $0$ ? That would give you the min and max, right? (note the above equation was found by simply substituting for $x$ )

One can easily find the derivative to be $-\frac{2 y^2}{\sqrt{1-y^2}}+2 \sqrt{1-y^2}-12 y$ , and you just need to set it equal to $0$ to find the $y$ values that give the local mins and maxes, then substitute the $y$ values into the above equation to get the values of the minimum and maximum.

(you do get $1+\sqrt{10}$ and $1-\sqrt{10}$ )

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1008 posts

kuing

#5 h Oct 23, 2010, 3:20 AM • 2 Y

Y by Adventure10, Mango247

another simpler solution

yezhu wrote:

$x^2 + 4y^2 = 4$ , let $f = x^2 + xy - 2y^2 = x^2 + px \cdot \frac{y}{p} - 2y^2 ,p \ne 0$ . then by Am-Gm, we have
$- \frac{1}{2}\left( {p^2 x^2 + \frac{{y^2 }}{{p^2 }}} \right) \le px \cdot \frac{y}{p} \le \frac{1}{2}\left( {p^2 x^2 + \frac{{y^2 }}{{p^2 }}} \right)$
$\Rightarrow - \frac{1}{2}\left( {\left( {p^2 - 2} \right)x^2 + \left( {\frac{1}{{p^2 }} + 4} \right)y^2 } \right) \le f \le \frac{1}{2}\left( {\left( {p^2 + 2} \right)x^2 + \left( {\frac{1}{{p^2 }} - 4} \right)y^2 } \right)$
(1), let $\frac{1}{{p^2 }} + 4 = 4\left( {p^2 - 2} \right) \Rightarrow p^2 - 2 = \frac{{\sqrt {10} - 1}}{2}$
$\Rightarrow f \ge - \frac{{\sqrt {10} - 1}}{4}\left( {x^2 + 4y^2 } \right) = 1 - \sqrt {10} \;;$
(2), let $\frac{1}{{p^2 }} - 4 = 4\left( {p^2 + 2} \right) \Rightarrow p^2 + 2 = \frac{{\sqrt {10} + 1}}{2}$
$\Rightarrow f \le \frac{{\sqrt {10} + 1}}{4}\left( {x^2 + 4y^2 } \right) = 1 + \sqrt {10} \;.$
...

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#6 h Oct 23, 2010, 12:21 PM • 1 Y

Y by Adventure10

zu allen. danke.

zum kuig. Ich verstehe nicht Ihre zweite Antwort.

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1008 posts

kuing

#7 h Oct 23, 2010, 12:33 PM • 1 Y

Y by Adventure10

Headhunter wrote:

zu allen. danke.

zum kuig. Ich verstehe nicht Ihre zweite Antwort.

I can't understand what you said, can you write it in english ?

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1581 posts

makar

#8 h Oct 23, 2010, 1:22 PM • 2 Y

Y by Adventure10, Mango247

Headhunter wrote:

Hello.

$x$ , $y$ are real numbers, where $x^{2}+4y^{2}=4$
Find Max (Min) of $x^{2}+xy-2y^{2}$ without trigonometry.

A much simpler solution

Let $x=2u\text{ and } y =v\implies u^2+v^2=1$

And the expression

$\begin{align*} x^{2}+xy-2y^{2}&=4u^2+2uv-2v^2\\ &=3(u^2-v^2)+2uv+1\end{align*}$

Now apply Cauchy Schwartz inequality on $3(u^2-v^2)+2uv$

$\begin{align*}|3\cdot (u^2-v^2)+1\cdot 2uv|&\le \sqrt{(3^2+1^2)\biggl((u^2-v^2)^2+(2uv)^2\biggr)}\\&=\sqrt{(10)\biggl((u^2+v^2)^2\biggr)}\\&=\sqrt{10}\end{align*}$

$\implies -\sqrt {10}\le 3(u^2-v^2)+ 2uv\le \sqrt{10}$

$\implies 1-\sqrt {10}\le 3(u^2-v^2)+ 2uv+1\le 1+\sqrt{10}$

$\implies 1-\sqrt {10}\le x^{2}+xy-2y^{2}\le 1+\sqrt{10}$

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SCP

#9 h Oct 23, 2010, 6:03 PM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

Headhunter wrote:

zu allen. danke.

zum kuig. Ich verstehe nicht Ihre zweite Antwort.

I can't understand what you said, can you write it in english ?

I don't know what language it is, but it is like :
to everyone, thanks.
to Kuing, I can't understand your ... answer.

I understand the solution, after you choose the values of $k$ .
So I assume it was a bit difficult to see for us how you knew to find the values.

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gaussintraining

1353 posts

gaussintraining

#10 h Oct 24, 2010, 4:04 AM • 2 Y

Y by Adventure10, Mango247

SCP wrote:

kuing wrote:

Headhunter wrote:

zu allen. danke.

zum kuig. Ich verstehe nicht Ihre zweite Antwort.

I can't understand what you said, can you write it in english ?

I don't know what language it is, but it is like :
to everyone, thanks.
to Kuing, I can't understand your ... answer.

It's German and (I believe) roughly translates to

"To all: Thanks."

"To kuing: I can't understand your second answer."

(Somebody correct me if I am wrong!)

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12750 posts

#11 h Oct 24, 2010, 11:08 AM • 1 Y

Y by Adventure10

Hi gaussintraining. You are correct.

This post has been edited 1 time. Last edited by AndrewTom, Oct 24, 2010, 11:22 AM

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#12 h Oct 24, 2010, 11:14 AM • 2 Y

Y by Adventure10, Mango247

to makar.
Your solution is really amazing.

to gaussintraining.
It's right translation.

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