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

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 $2x+3y=\boxed{\sqrt{26}}$, 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 $\frac{4}{x^2}=\frac{9}{y^2}\implies 2y=3x$.

Third Solution

Imagine the equations graphed in the coordinate plane. $x^2+y^2=2$ is a circle centered at the origin with

radius $\sqrt{2}$. $2x+3y=k$ is a line. We want to find the largest value of $k$ such that the line

intersects the circle, giving real number solutions for $x$ and $y$. This occurs when $2x+3y=k$ is tangent

to the circle, and thus when the distance from the line to the origin is $\sqrt{2}$. The distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$ is


$\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$.


Plugging in $A=2, B=3, C=-k, x_0=0$, and $y_0=0$ and setting the expression equal to $\sqrt{2}$ yields

$\pm\frac{k}{\sqrt{13}}=\sqrt{2}$, or $k=\pm\sqrt{26}$. We want the largest value of $k$, so

$k=\boxed{\sqrt{26}}$ is the highest possible value.

Fourth Solution

By AM-QM, $\frac{2x + 3y}{13} = \frac{4 \cdot \frac{x}{2} + 9 \cdot \frac{y}{3}}{13} \le \sqrt{\frac{4 \cdot \left(\frac{x}{2}\right)^2 + 9 \left(\frac{y}{3}\right)^2}{13}} = \sqrt{\frac{2}{13}}$, so $2x + 3y \le \sqrt{26}$, equality when $\frac x2 = \frac y3$.

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