f ' (x) = ax^(a-1)(1-x)^b + b(1-x)^(b-1) x^a
= 0 for a max of f(x)
x^(a-1) (1-x)^(b-1)[a(1-x) + b(x)] = 0
then x^(a-1) = 0 , which is not possible
or (1-x)^(b-1) = 0 , which is also not possible
or
a-ax+bx=0
a = ax+bx
a = x(a+b)
x = a/(a+b)
f(a/(a+b))) = [a/(a+b)]^a [(1-a/(a+b)]^b
= [a/(a+b)]^a [b/(a+b)]^b
the max is [a/(a+b)]^a [b/(a+b)]^b
suppose we let a= 2 and b = 3
f(x) = x^2(1-x)^3 and graph it
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%5E2%281-x%29%5E3
the max should occur at x = 2/5 or .4 , which it is
and it should be (2/5)^2 (3/5)^3 = appr .035 , which it is
It is extremely likely that my answer is correct
If a and b are positive numbers, find the maximum value of f(x)=x^a(1−x)^b, 0≤x≤1
Your answer may depend on a and b.
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2 answers