Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 1"
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<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)=(\dfrac{(a-1)^3}{a^3}+1)(\dfrac{1}{a^3}+1)=\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}</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> |
As a gets large, the fraction gets small, therefore maximizing <math>(x^3+1)(y^3+1)</math>. But when a gets small(up to 2), the fraction gets bigger, and therefore lessens <math>(x^3+1)(y^3+1)</math>. | As a gets large, the fraction gets small, therefore maximizing <math>(x^3+1)(y^3+1)</math>. But when a gets small(up to 2), the fraction gets bigger, and therefore lessens <math>(x^3+1)(y^3+1)</math>. |
Revision as of 10:45, 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.