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## Problem:

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**.

## Chatuspathi

**What is this topic:**Real Analysis**What are some of the associated concept:**Supremum Property, Maxima-Minima**Book Suggestions:**Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert