Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 1"
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''Dan Schwarz'' | ''Dan Schwarz'' | ||
==Solution== | ==Solution== | ||
| + | If y is negative, then <math>(x^3+1)(y^3+1)</math> is also negative, so we want <math>0\leq y\leq 1</math>. | ||
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| + | <math>y=\dfrac{1}{a}</math> | ||
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| + | where <math>a<0\leq 1</math>. Let's see what happens when a gets large: | ||
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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> | ||
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| + | <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> | ||
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| + | 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>. | ||
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| + | Therefore, the maximum value of <math>(x^3+1)(y^3+1)</math> is when x=1 and y=0, which is 2. | ||
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==See also== | ==See also== | ||
*[[2006 Romanian NMO Problems]] | *[[2006 Romanian NMO Problems]] | ||
[[Category: Olympiad Algebra Problems]] | [[Category: Olympiad Algebra Problems]] | ||
Revision as of 11:44, 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:
$=\dfrac{a^6-3a^5+4a^4-3a^2+4a-1}{a^6}=1-\dfrac{3a^5-4a^4+3a^2-4a+1}$ (Error compiling LaTeX. Unknown error_msg)
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.
