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AoPS Wiki talk:Problem of the Day/June 21, 2011

Revision as of 20:37, 20 June 2011 by Btilm305 (talk | contribs) (latex errors)

Problem

AoPSWiki:Problem of the Day/June 21, 2011

Solution

$(2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26$. Hence $(2x+3y)^2 \le 26$ or $2x+3y = \sqrt{26}$. If $3x-2y=0$ and $2x+3y=\sqrt{26}$, then $2x+3y$ attains this maximum value on the circle $x^2+y^2=2$.

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