## anonymous 4 years ago If a and b are positive numbers, find the maximum value of f(x)=x^a*(1-x)^b, 0 less than or equal to x less than or equal to 1. Your answer may depend on a and b. What is the maximum value?

1. amistre64

maximum value of: $f(x)=x^a*(1-x)^b;\ 0\le x \le 1$

2. amistre64

$f'(x)=ax^{(a-1)}(1-x)^b-x^ab(1-x)^{(b-1)}$ $ax^{(a-1)}(1-x)^b-x^ab(1-x)^{(b-1)}=0$ solving this might get us some critical points

3. anonymous

Critical point = a/(a+b)?

4. anonymous

Got the answer. It is (a/(a+b))^a(b/(a+b))^b

5. anonymous

Thanks for helping

6. amistre64

$ax^{(a-1)}(1-x)^b=x^ab(1-x)^{(b-1)}$ $ax^{(a-1)}x^{-a}(1-x)^b=b(1-x)^{(b-1)}$ $ax^{-1}(1-x)^b=b(1-x)^{(b-1)}$ $a=xb(1-x)^{(b-1)}(1-x)^{-b}$ $a=xb(1-x)^{-1}$ $\frac{a}{b}=\frac{x}{1-x}$ $a-ax=bx$ $a=bx+ax$ $\frac{a}{b+a}=x$ looks to be that way if i did it right