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Difference between revisions of "2006 Romanian NMO Problems/Grade 9/Problem 1"
(→‎Solution)
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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>a<0\leq 1</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>=\dfrac{a^6-3a^5+4a^4-3a^2+4a-1}{a^6}=1-\dfrac{3a^5-4a^4+3a^2-4a+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>.
 +
 +
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]]
 
[[Category: Olympiad Algebra Problems]]

Revision as of 10:44, 10 October 2007

Problem

Find the maximal value of

$\left( x^3+1 \right) \left( y^3 + 1\right)$,

where $x,y \in \mathbb R$, $x+y=1$.

Dan Schwarz

Solution

If y is negative, then $(x^3+1)(y^3+1)$ is also negative, so we want $0\leq y\leq 1$.

$y=\dfrac{1}{a}$

where $a<0\leq 1$. Let's see what happens when a gets large:

$(x^3+1)(y^3+1)=(\dfrac{(a-1)^3}{a^3}+1)(\dfrac{1}{a^3}+1)=\dfrac{(a^3+1)((a-1)^3+a)}{a^6}$

$=\dfrac{a^6-3a^5+4a^4-3a^2+4a-1}{a^6}=1-\dfrac{3a^5-4a^4+3a^2-4a+1}$ (Error compiling LaTeX. Unknown error_msg)

As a gets large, the fraction gets small, therefore maximizing $(x^3+1)(y^3+1)$. But when a gets small(up to 2), the fraction gets bigger, and therefore lessens $(x^3+1)(y^3+1)$.

Therefore, the maximum value of $(x^3+1)(y^3+1)$ is when x=1 and y=0, which is 2.

See also

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