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