Find maximum of a polynomial? How would I find the value of p which the function f is the maximum:
$f(p)= 100 p(1-p)^{99}$  $p\geq0$
And in general, can setting the derivative equal to $0$ and solving for the variable lead to the values in function's domain that produces the maximum value ?
 A: In short, yes. This is known as the "Fermat's Theorem". Of course, you have to check with the second derivative test to see if you found a maximum or minimum. You could also get a saddle point, which is effectively a place where the graph is flat... see here.  
Let's find the maximum of your function:
$$f(p)= 100 p(1-p)^{99}$$
$$f'(p) = 100 (1-p)^{99}-9900 (1-p)^{98} p$$
Now set the derivative to $0$:
$$0 = 100 (1-p)^{99}-9900 (1-p)^{98} p$$
$$0=-100(p-1)^{98}(100p-1)$$
$$(p-1)^{98} =0 \;\text{ or } \;100p-1 = 0$$
$$p = 1 \;\text{ or } \;p = \frac{1}{100}$$
After some checking we find the second solution is the global maximum, giving a value of $\approx 0.36973$. Thus, we find that the maximum is around the point $(0.01,0.36973)$
A: Here is another way;  clearly we seek $p \in (0, 1)$, so $p, 1-p$ are both positive.
$$\frac{99}{100}f(p) = (99p) \cdot \underbrace{(1-p)(1-p)\cdots(1-p)}_{99 \text{ times}} $$
Now the RHS is the product of $100$ terms, which sum to a constant, viz. $99$.  Therefore its maximum is when all terms are equal, viz. $\dfrac{99}{100}$ (this can be shown easily with AM-GM).  $\implies p = \dfrac1{100}$.
