Answer
The maximum value of $f$ on this interval is $f(\frac{a}{a+b}) = \frac{a^a~b^b}{(a+b)^{a+b}}$
Work Step by Step
We can find the value of $x$ when $f'(x) = 0$:
$f(x) = x^a(1-x)^b$
$f'(x) = ax^{a-1}(1-x)^b+b(1-x)^{b-1}(-1)x^a = 0$
$ax^{a-1}(1-x)^b-b(1-x)^{b-1}x^a = 0$
$ax^{a-1}(1-x)^b = b(1-x)^{b-1}x^a$
$a(1-x) = bx$
$a-ax = bx$
$a = ax+bx$
$x = \frac{a}{a+b}$
We can find the value of $f(\frac{a}{a+b})$:
$f(x) = x^a(1-x)^b$
$f(\frac{a}{a+b}) = (\frac{a}{a+b})^a(1-\frac{a}{a+b})^b$
$f(\frac{a}{a+b}) = \frac{a^a}{(a+b)^a}\cdot (\frac{a+b}{a+b}-\frac{a}{a+b})^b$
$f(\frac{a}{a+b}) = \frac{a^a}{(a+b)^a}\cdot (\frac{b}{a+b})^b$
$f(\frac{a}{a+b}) = \frac{a^a}{(a+b)^a}\cdot \frac{b^b}{(a+b)^b}$
$f(\frac{a}{a+b}) = \frac{a^ab^b}{(a+b)^{a+b}}$
Since $a$ and $b$ are both positive, $f(\frac{a}{a+b}) \gt 0$
Note that $f(0) = f(1) = 0$
Therefore, the maximum value of $f$ on this interval is $f(\frac{a}{a+b}) = \frac{a^a~b^b}{(a+b)^{a+b}}$
