# Stumped on Maximum value Problem

• Nov 19th 2009, 03:22 PM
Stumped on Maximum value Problem
I want to find the maximum value of $\displaystyle f(x)=x^a(1-x)^b$ on $\displaystyle 0\leq x \leq 1$ and $\displaystyle a,b>0$.

I just apply the standard procedure for find the extrema of a function:

$\displaystyle f'(x)=ax^{a-1}(1-x)^b-b(1-x)^{b-1}$

$\displaystyle ax^{a-1}(1-x)^b-b(1-x)^{b-1}=0$

$\displaystyle ax^{a-1}(1-x)^b=b(1-x)^{b-1}$

I Divide both sides by $\displaystyle (1-x)^b$ to obtain:

$\displaystyle ax^{a-1}=\frac{b}{1-x}$

This doesn't really seem to be going anywhere. I haven't been able to isolate x to obtain a function value. Any tips?
• Nov 19th 2009, 03:30 PM
Jester
Quote:

I want to find the maximum value of $\displaystyle f(x)=x^a(1-x)^b$ on $\displaystyle 0\leq x \leq 1$ and $\displaystyle a,b>0$.

I just apply the standard procedure for find the extrema of a function:

$\displaystyle f'(x)=ax^{a-1}(1-x)^b-b {\color{red} x^a} (1-x)^{b-1}$

$\displaystyle ax^{a-1}(1-x)^b-b(1-x)^{b-1}=0$

$\displaystyle ax^{a-1}(1-x)^b=b(1-x)^{b-1}$

I Divide both sides by $\displaystyle (1-x)^b$ to obtain:

$\displaystyle ax^{a-1}=\frac{b}{1-x}$

This doesn't really seem to be going anywhere. I haven't been able to isolate x to obtain a function value. Any tips?

I believe you've forgot a term (see above in red).
• Nov 19th 2009, 04:02 PM