Positive number whose square exceeds its cube by the greatest amount 
Find the positive number whose square exceeds its cube by the greatest amount.

Could you please help me with this question? It seems pretty easy at first glance, but I struggled to produce any answer. Thank you in advance!
UPD:
    Sorry, I didn't mention, but the exercise is to solve it without any differentiation.
 A: Let x be the number. We're trying to find the number $x$ where $x^2 > x^3$, and this is the biggest possible. So we're trying to find the $x$ that maximizes $x^2 - x^3$ We want $x^2 - x^3$ to be a maximum. 
Differentiate we get $2x - 3x^2$ and
set to zero, we find maximum at $x_1 = 0, x_2= \frac{2}{3}$ therefore $x=\frac{2}{3}$ is the answer. 
A: Your problem is naturally expressed as a constrained optimisation problem:
$$ \max_{x} (x^2 - x^3) \quad \text{s.t} \quad x \geq 0 $$
It is also easy to see that the solution must be below 1.
Now you could go and use some fancy optimisation technique, such as formulating the Lagrangian, finding the critical points, verifying the KKT conditions with the constraint active... But here it's pretty much overkill.
Let $f(x) = x^2 - x^3$, then $f'(x) = 2x - 3x^2$. Every optimum is a critical point, so the optimal $x$ satisfies $2x = 3x^2$.
We want positive numbers, if $x = 0$ then $x^2 = x^3 = 0$. If $x \neq 0$ then $2 = 3x$ and $x = \frac{2}{3}$. For $x = \frac{2}{3}, \; x^2 - x^3 = \frac{4}{9} - \frac{8}{27} = \frac{4}{27}$.
So the solution is $x = \frac{2}{3}$.
A: y = x^2 - x^3
We want the maximum point of the curve (where the square exceeds the cube by the highest amount) so we differentiate.
dy/dx = 2x - 3x^2
The maximum point is when dy/dx = 0 and d^2y/dx^2 (the second derivative) < 0
2x = 3x^2
x = 0 or 2 = 3x
x = 0 or 2/3
d^2y/dx^2 = 2 - 6x
Substitute in 0: d^2y/dx^2 = 2 so invalid value as d^2y/dx^2 should be < 0
Substitute in 2/3: d^2y/dx^2 = -2 so the answer is 2/3
A: Here is a method that does not require calculus:


Lemma: If $x+y+z = k$ (where $k$ is a constant), the product $xyz$ is maximized when $x=y=z= \dfrac k3 $

Proof:
By $AM-GM$, we have 
$$\dfrac {x+y+z}{3} \ge \sqrt[3] {xyz}$$
$$\dfrac {k}{3} \ge \sqrt[3] {xyz}$$
$$\left(\dfrac {k}{3}\right)^3 \ge  xyz$$
We can see that the maximum of $xyz$ is $\left(\dfrac {k}{3}\right)^3$, with equality occurring $\iff x=y=z$
$$\left(\dfrac {k}{3}\right)^3 = x\cdot x \cdot x$$
$\implies x = \dfrac k3 $
$\implies x = y = z = \dfrac k3$
Remark: This lemma can be generalized for any $n$.

Now we want to find the maximum of $x^2 - x^3$, or equivalently 
$$ f(x) = x\cdot x \cdot (1-x)$$
The maximum of $f$ occurs at the same $x$ value as the maximum of 
$$g(x) = 2\cdot f(x) = x \cdot x \cdot (2-2x)$$
But notice that the factors in $g$ have a constant sum:
$$(x) + (x) + (2-2x) = 2$$
By the lemma, the product $x \cdot x \cdot (2-2x)$ reaches its maximum when $x = x = (2-2x) = \dfrac 23$
We can see that $x = \dfrac 23$, and the maximum is $$\left(\dfrac 23 \right )^2 - \left( \dfrac 23 \right)^3 = \dfrac {4}{27}$$
