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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.

 

October 27, 2017

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