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Numerade Educator



Problem 63 Hard Difficulty

If $a$ and $b$ are positive numbers, find the maximum value of $ f(x) = x^a (1 - x)^b $, $ 0 \leqslant x \leqslant 1 $.


\[\begin{aligned}f(x) &=x^{a}(1-x)^{b}, 0 \leq x \leq 1, a>0, b>0 \\
f^{\prime}(x) &=x^{a} \cdot b(1-x)^{b-1}(-1)+(1-x)^{b} \cdot a x^{a-1}=x^{a-1}(1-x)^{b-1}[x \cdot b(-1)+(1-x) \cdot a] \\
&=x^{a-1}(1-x)^{b-1}(a-a x-b x)\end{aligned}
\]At the endpoints, we have $f(0)=f(1)=0 \quad[\text { the minimum value of } f] .$ In the interval $(0,1), f^{\prime}(x)=0 \Leftrightarrow x=\frac{a}{a+b}$\[
f\left(\frac{a}{a+b}\right)=\left(\frac{a}{a+b}\right)^{a}\left(1-\frac{a}{a+b}\right)^{b}=\frac{a^{a}}{(a+b)^{a}}\left(\frac{a+b-a}{a+b}\right)^{b}=\frac{a^{a}}{(a+b)^{a}} \cdot \frac{b^{b}}{(a+b)^{b}}=\frac{a^{a} b^{b}}{(a+b)^{a+b}}
\]So $f\left(\frac{a}{a+b}\right)=\frac{a^{a} b^{b}}{(a+b)^{a+b}}$ is the absolute maximum value

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