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I need some help with the following problem.

How, I can find the, Maximum Value of the function $f(x)=x^n(1-x)^n$ for a natural number $n \geq1$ and $x\in[0,1]$.

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thanks !! sir !! –  ram Dec 31 '12 at 10:55

6 Answers 6

up vote 12 down vote accepted

Note that $$x^n(1-x)^n=\bigl(x(1-x)\bigr)^n=\left(\frac14-\left(\frac12-x\right)^2\right)^n\le \frac1{4^n}$$

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First, find the maximum of $x(1-x)$, and let it have value $Q$. You can do this by completing the square, AM-GM, or differentiation (depending on how much you know).

Then since the exponentiation function $x \rightarrow x^n$ is a strictly increasing function for $x\in [0,\infty]$ and $n\geq 1$, if follows that $Q^n$ will be the maximum value of your function.

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Since your function is positive you can take the $n^{th}$ root and maximise that, and then raise that value to the $n^{th}$ power.

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$$f(x)=x^n(1-x)^n$$ $$f'(x)=n(x-x^2)^{n-1}(1-2x)=0$$ $$x-x^2=x(1-x)=0,x=0,x=1$$ and $f(0)=f(1)=0$ or $$1-2x=0,x=1/2$$ so maximum is for $x=1/2$ $$f(1/2)=(1/2)^{2n}$$

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Since your function is positive, you can take the logarithm and maximize it; in the end you can exponentiate the result. In formulas $$ \ln f(x) = n [\log x + \log (1-x)].$$ Then $$\frac{d}{dx} \ln f(x^*) = \frac{n}{x^*} - \frac{n}{1-x^*} =0$$ leads to $x^*=1/2$. You then have $f(x^*) = 1/2^{2n}$ which you should still compare with the values at $x=0,1$.

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Hint: Note that $f(x)$ is the product of $2n$ nonnegative numbers. Now you may use A.M.$\ge$ G.M.

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