Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

Revision as of 09:28, 21 June 2011 by Tonypr (talk | contribs)

Problem

AoPSWiki:Problem of the Day/June 21, 2011

Solutions

First 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 $x^2+y^2=2$.

Second Solution

Let $x$ and $y$ be real numbers such that $x^2+y^2=2$. Note that \[|x|^2+|y|^2=2 \text{ and } 2|x|+3|y|\ge 2x+3y\] thus, we may assume that $x$ and $y$ are positive. Furthermore, by the Cauchy-Schwarz Inequality, we have \[(4+9)(x^2+y^2)\ge (2x+3y)^2\] but since $x^2+y^2=2$, the inequality is equivalent with \[26\ge (2x+3y)^2\] or \[\sqrt{26}\ge 2x+3y\] so the maximum is $\boxed{\sqrt{26}}$ and it is reached when $\frac{4}{x^2}=\frac{9}{y^2}\implies 2y=3x$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=AoPS_Wiki_talk:Problem_of_the_Day/June_21,_2011&oldid=39845"