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>. | ||
+ | |||
+ | <math>y=\dfrac{1}{a}</math> | ||
+ | |||
+ | where <math>1 \leq a</math>. Let's see what happens when a gets large: | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | Therefore, the maximum value of <math>(x^3+1)(y^3+1)</math> is when x=1 and y=0, which is 2. | ||
+ | |||
==See also== | ==See also== | ||
*[[2006 Romanian NMO Problems]] | *[[2006 Romanian NMO Problems]] | ||
+ | [[Category: Olympiad Algebra Problems]] |
Latest revision as of 21:36, 10 October 2007
Problem
Find the maximal value of
![$\left( x^3+1 \right) \left( y^3 + 1\right)$](http://latex.artofproblemsolving.com/5/c/9/5c9771df93a5ab9e4e2c14fbf59265a888c74397.png)
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.