Thread: Stumped on Maximum value Problem

1. 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?

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).

3. Originally Posted by Danny
I believe you've forgot a term (see above in red).
thanks, I had a feeling this should be an easy one. I was able to isolate x after fixing the error.