💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

Numerade Educator

Like

Report

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 $.

Answer

\[\begin{aligned}f(x) &=x^{a}(1-x)^{b}, 0 \leq x \leq 1, a>0, b>0 \\
f^{\prime}(x) &=x^{a} \cdot b(1-x)^{b-1}(-1)+(1-x)^{b} \cdot a x^{a-1}=x^{a-1}(1-x)^{b-1}[x \cdot b(-1)+(1-x) \cdot a] \\
&=x^{a-1}(1-x)^{b-1}(a-a x-b x)\end{aligned}
\]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

More Answers

Discussion

You must be signed in to discuss.

Video Transcript

using the product Jewel and powerful. We know we can take the derivative we end up with X zero arcs is Juan and axes a over a plus B. These are the places where the driven of zero no, you know that awful 00 off of one is zero and us of a over a plus B was eight of the eighth B to the B over a plus B. The power of a post be the absolute maximum is a over a plus. B is off of a over a plus B, which is eight of the a times B the beat over a plus B the power of a plus b So this is the absolute max.