Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 21, 2011"
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==Problem== | ==Problem== | ||
{{:AoPSWiki:Problem of the Day/June 21, 2011}} | {{:AoPSWiki:Problem of the Day/June 21, 2011}} | ||
− | ==Solution== | + | ==Solutions== |
− | (2x+3y)^2 + (3x-2y)^2 = 13(x^2+y^2)=26 | + | === First Solution === |
− | Hence (2x+3y)^2 < | + | <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>. |
− | If 3x-2y=0 and 2x+3y=sqrt | + | If <math>3x-2y=0</math> and <math>2x+3y=\boxed{\sqrt{26}}</math>, then <math>2x+3y</math> attains this maximum value on the circle <math>x^2+y^2=2</math>. |
+ | === Second Solution === | ||
+ | Let <math>x</math> and <math>y</math> be real numbers such that <math>x^2+y^2=2</math>. Note that | ||
+ | <cmath>|x|^2+|y|^2=2 \text{ and } 2|x|+3|y|\ge 2x+3y</cmath> | ||
+ | thus, we may assume that <math>x</math> and <math>y</math> are positive. Furthermore, by the [[Cauchy-Schwarz Inequality]], we have | ||
+ | <cmath>(4+9)(x^2+y^2)\ge (2x+3y)^2</cmath> | ||
+ | but since <math>x^2+y^2=2</math>, the inequality is equivalent with | ||
+ | <cmath>26\ge (2x+3y)^2</cmath> | ||
+ | or | ||
+ | <cmath>\sqrt{26}\ge 2x+3y</cmath> | ||
+ | so the maximum is <math>\boxed{\sqrt{26}}</math> and it is reached when <math>\frac{4}{x^2}=\frac{9}{y^2}\implies 2y=3x</math>. | ||
+ | ===Third Solution=== | ||
+ | Imagine the equations graphed in the coordinate plane. <math> x^2+y^2=2 </math> is a circle centered at the origin with | ||
+ | |||
+ | radius <math> \sqrt{2} </math>. <math> 2x+3y=k </math> is a line. We want to find the largest value of <math> k </math> such that the line | ||
+ | |||
+ | intersects the circle, giving real number solutions for <math> x </math> and <math> y </math>. This occurs when <math> 2x+3y=k </math> is tangent | ||
+ | |||
+ | to the circle, and thus when the distance from the line to the origin is <math> \sqrt{2} </math>. The distance from a point <math> | ||
+ | |||
+ | (x_0, y_0) </math> to a line <math> Ax+By+C=0 </math> is | ||
+ | |||
+ | |||
+ | |||
+ | <math> \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} </math>. | ||
+ | |||
+ | |||
+ | |||
+ | Plugging in <math> A=2, B=3, C=-k, x_0=0 </math>, and <math> y_0=0 </math> and setting the expression equal to <math> \sqrt{2} </math> yields | ||
+ | |||
+ | <math> \pm\frac{k}{\sqrt{13}}=\sqrt{2} </math>, or <math> k=\pm\sqrt{26} </math>. We want the largest value of <math> k </math>, so | ||
+ | |||
+ | <math> k=\boxed{\sqrt{26}} </math> is the highest possible value. | ||
+ | |||
+ | === Fourth Solution === | ||
+ | By [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality|AM-QM]], <math>\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}}</math>, so <math>2x + 3y \le \sqrt{26}</math>, equality when <math>\frac x2 = \frac y3</math>. |
Latest revision as of 16:14, 22 June 2011
Contents
Problem
AoPSWiki:Problem of the Day/June 21, 2011
Solutions
First Solution
. Hence or . If and , then attains this maximum value on the circle .
Second Solution
Let and be real numbers such that . Note that thus, we may assume that and are positive. Furthermore, by the Cauchy-Schwarz Inequality, we have but since , the inequality is equivalent with or so the maximum is and it is reached when .
Third Solution
Imagine the equations graphed in the coordinate plane. is a circle centered at the origin with
radius . is a line. We want to find the largest value of such that the line
intersects the circle, giving real number solutions for and . This occurs when is tangent
to the circle, and thus when the distance from the line to the origin is . The distance from a point to a line is
.
Plugging in , and and setting the expression equal to yields
, or . We want the largest value of , so
is the highest possible value.
Fourth Solution
By AM-QM, , so , equality when .