Stumped on Maximum value Problem

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November 19th 2009, 03:22 PM
adkinsjr

Stumped on Maximum value Problem

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

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

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

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

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

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

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

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

Quote:

Originally Posted by adkinsjr

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

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

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

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

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

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

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

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).
November 19th 2009, 04:02 PM
adkinsjr

Quote:

Originally Posted by Danny

I believe you've forgot a term (see above in red).

(Giggle) thanks, I had a feeling this should be an easy one. I was able to isolate x after fixing the error.

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