<s>[INST] 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.
 [/INST] 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 $$\left(\dfrac 23 \right )^2 - \left( \dfrac 23 \right)^3 = \dfrac {4}{27}$$
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