Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 21, 2011"
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<math> k=\boxed{\sqrt{26}} </math> is the highest possible value. | <math> k=\boxed{\sqrt{26}} </math> is the highest possible value. | ||
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+ | === 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 .