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?

Question

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

Critical point = a/(a+b)?

anonymous · Answer

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

anonymous · Answer

Thanks for helping

amistre64 · Answer

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