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moneybird
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?
maximum value of: \[f(x)=x^a*(1-x)^b;\ 0\le x \le 1\]
\[f'(x)=ax^{(a-1)}(1-x)^b-x^ab(1-x)^{(b-1)}\] \[ax^{(a-1)}(1-x)^b-x^ab(1-x)^{(b-1)}=0\] solving this might get us some critical points
Critical point = a/(a+b)?
Got the answer. It is (a/(a+b))^a(b/(a+b))^b
\[ax^{(a-1)}(1-x)^b=x^ab(1-x)^{(b-1)}\] \[ax^{(a-1)}x^{-a}(1-x)^b=b(1-x)^{(b-1)}\] \[ax^{-1}(1-x)^b=b(1-x)^{(b-1)}\] \[a=xb(1-x)^{(b-1)}(1-x)^{-b}\] \[a=xb(1-x)^{-1}\] \[\frac{a}{b}=\frac{x}{1-x}\] \[a-ax=bx\] \[a=bx+ax\] \[\frac{a}{b+a}=x\] looks to be that way if i did it right