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TIFR 2014 Problem 13 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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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