Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 21, 2011"
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==Solution== | ==Solution== | ||
| − | (2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26 | + | <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>. |
| − | Hence (2x+3y)^2 < | + | 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 3x-2y=0 and 2x+3y=sqrt | ||
Revision as of 20:37, 20 June 2011
Problem
AoPSWiki:Problem of the Day/June 21, 2011
Solution
. Hence 
or 
.
If 
and 
, then 
attains this maximum value on the circle 
.
