Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 21, 2011"
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==Problem== | ==Problem== | ||
{{:AoPSWiki:Problem of the Day/June 21, 2011}} | {{:AoPSWiki:Problem of the Day/June 21, 2011}} | ||
| − | ==Solution== | + | ==Solutions== |
| + | === First Solution === | ||
<math>(2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26</math>. Hence <math>(2x+3y)^2 \le 26</math> or <math>2x+3y = \sqrt{26}</math>. | <math>(2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26</math>. Hence <math>(2x+3y)^2 \le 26</math> or <math>2x+3y = \sqrt{26}</math>. | ||
If <math>3x-2y=0</math> and <math>2x+3y=\sqrt{26}</math>, then <math>2x+3y</math> attains this maximum value on the circle <math>x^2+y^2=2</math>. | If <math>3x-2y=0</math> and <math>2x+3y=\sqrt{26}</math>, then <math>2x+3y</math> attains this maximum value on the circle <math>x^2+y^2=2</math>. | ||
| + | === Second Solution === | ||
| + | Let <math>x</math> and <math>y</math> be real numbers such that <math>x^2+y^2=2</math>. Note that | ||
| + | <cmath>|x|^2+|y|^2=2 \text{ and } 2|x|+3|y|\ge 2x+3y</cmath> | ||
| + | thus, we may assume that <math>x</math> and <math>y</math> are positive. Furthermore, by the [[Cauchy-Schwarz Inequality]], we have | ||
| + | <cmath>(4+9)(x^2+y^2)\ge (2x+3y)^2</cmath> | ||
| + | but since <math>x^2+y^2=2</math>, the inequality is equivalent with | ||
| + | <cmath>26\ge (2x+3y)^2</cmath> | ||
| + | or | ||
| + | <cmath>\sqrt{26}\ge 2x+3y</cmath> | ||
| + | so the maximum is <math>\boxed{\sqrt{26}}</math> and it is reached when <math>\frac{4}{x^2}=\frac{9}{y^2}\implies 2y=3x</math>. | ||
Revision as of 10:28, 21 June 2011
Problem
AoPSWiki:Problem of the Day/June 21, 2011
Solutions
First Solution
. Hence 
or 
.
If 
and 
, then 
attains this maximum value on the circle 
.
Second Solution
Let 
and 
be real numbers such that . Note that
![\[|x|^2+|y|^2=2 \text{ and } 2|x|+3|y|\ge 2x+3y\]](http://latex.artofproblemsolving.com/0/d/a/0da04f48a1f27299e8338dda1b646941e5edaf40.png)
thus, we may assume that and are positive. Furthermore, by the Cauchy-Schwarz Inequality, we have
![\[(4+9)(x^2+y^2)\ge (2x+3y)^2\]](http://latex.artofproblemsolving.com/e/3/6/e36fc1ba47d25ab9ae5f72fc4a519be0acaea6c5.png)
but since , the inequality is equivalent with
![\[26\ge (2x+3y)^2\]](http://latex.artofproblemsolving.com/c/a/0/ca07ba3447d9b2423d3ec83e515f47628267adb9.png)
or
![\[\sqrt{26}\ge 2x+3y\]](http://latex.artofproblemsolving.com/6/e/e/6eeb459243ea88d0cda1ad610deead540c23cef8.png)
so the maximum is 
and it is reached when 
.
