Like
Report
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 $.
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.
Top Calculus 2 / BC Educators
Kristen K.
University of Michigan - Ann Arbor
Michael J.
Idaho State University
Joseph L.
Boston College
Watch More Solved Questions in Chapter 4
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.
Top Calculus 2 / BC Educators
Grace H.
Numerade Educator
Kristen K.
University of Michigan - Ann Arbor
Michael J.
Idaho State University
Joseph L.
Boston College
