Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 1"
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<math>y=\dfrac{1}{a}</math> | <math>y=\dfrac{1}{a}</math> | ||
| − | where <math> | + | where <math>1 \leq a</math>. Let's see what happens when a gets large: |
| − | <math>(x^3+1)(y^3+1)=(\dfrac{(a-1)^3}{a^3}+1)(\dfrac{1}{a^3}+1)=\dfrac{(a^3+1)((a-1)^3+a)}{a^6}</math> | + | <math>(x^3+1)(y^3+1)=\left(\dfrac{(a-1)^3}{a^3}+1\right)\left(\dfrac{1}{a^3}+1\right)=\dfrac{(a^3+1)((a-1)^3+a)}{a^6}</math> |
<math>=\dfrac{a^6-3a^5+4a^4-3a^2+4a-1}{a^6}=1-\dfrac{3a^5-4a^4+3a^2-4a+1}{a^6}</math> | <math>=\dfrac{a^6-3a^5+4a^4-3a^2+4a-1}{a^6}=1-\dfrac{3a^5-4a^4+3a^2-4a+1}{a^6}</math> | ||
Latest revision as of 20:36, 10 October 2007
Problem
Find the maximal value of
,where 
, 
.
Dan Schwarz
Solution
If y is negative, then 
is also negative, so we want 
.
where 
. Let's see what happens when a gets large:
As a gets large, the fraction gets small, therefore maximizing . But when a gets small(up to 2), the fraction gets bigger, and therefore lessens .
Therefore, the maximum value of is when x=1 and y=0, which is 2.
