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# Supremum of function (TIFR 2014 problem 13)

Question:

Let $S$ be the set of all tuples $(x,y)$ with $x,y$ non-negative real numbers satisfying $x+y=2n$ ,for a fixed $n\in\mathbb{N}$. Then the supremum value of $x^2y^2(x^2+y^2)$ on the set $S$ is:

A. $3n^6$

B. $2n^6$

C. $4n^6$

D. $n^6$

Discussion:

Write the expression in terms of $x$ only by substituting $y=2n-x$.

Let $f(x)=x^2(2n-x)^2(x^2+(2n-x)^2)$. Here, $x\in[0,2n]$.

Note that for $x=0$ or $x=2n$ the function $f(x)=0$. Also, $f$ is positive everywhere else on the interval.

So we want to find $sup\{f(x)|x\in (0,2n)\}$. Note that it exists because the interval is compact and $f$ is continuous.

One can straightaway take derivative and compute, or one can do the following:

Take log. Note that now we are only working on the open interval $(0,2n)$.

$log(f(x))=2logx+2log(2n-x)+log(x^2+(2n-x)^2)$

Now take derivative.

$\frac{f'(x)}{f(x)}=\frac{2}{x}+\frac{-2}{2n-x}+\frac{2x-2(2n-x)}{x^2+(2n-x)^2}$

$=\frac{4n-4x}{x(2n-x)}+\frac{4n-4x}{x^2+(2n-x)^2}$

$=4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]$.

Now, $f'(x)=0$ if and only if $4(n-x)[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]=0$.

Note that $[\frac{1}{x(2n-x)}+\frac{1}{x^2+(2n-x)^2}]>0$ for all $x\in (0,2n)$.

Therefore, $f'(x)=0$ if and only if $x=n$. If now, $x$ is slightly bigger than $n$ then $\frac{f'(x)}{f(x)}<0$ and since $f(x)>0$ we have $f'(x)<0$ in that case. And if $x$ is slightly smaller than $n$ then $f'(x)>0$.

This proves that indeed the point $x=n$ is a point of maxima.

Therefore, the supremum value is $f(n)=2n^6$. So the correct answer is option B.

## By Ashani Dasgupta

Ph.D. in Mathematics from University of Wisconsin Milwaukee (USA)
Founder - Faculty at Cheenta