<s>[INST] Rectangle of maximum area under the curve $y=x(x-1)^2$ A rectangle's bottom is $y=0$. While the top corners are on the curve $y=x(x-1)^2$ between $x=0$ and $x=1$. Find the maximum area of this rectangle.
My Progress
Defining $f(x) =x(x-1) ^2$
I first starting by assuming that there exist $a$ and $b$ such that $0<a, b<1$ and $f(a) =f(b)$.
The area of this rectangle is given by $|b-a|f(a) $.
We know that $a(a-1) ^2=b(b-1) ^2$. Neglecting the $a=b$ case we get that $$a^2+b^2+ab-2(a+b) +1=0$$
Now it remains to maximise $$|b-a|a(a-1) ^2$$ under the constraint $$a^2+b^2+ab-2(a+b) +1=0$$
But now I'm pretty much stuck here.
 [/INST] There is really only one free variable, which is without loss of generality, the $x$ coordinate of the vertex of the rectangle that is closest to the origin.  If we call this value $a$, then the other vertex coordinates are uniquely defined:  in total, we have
$$(a,0), (a, a(1-a)^2), (b,a(1-a)^2), (b, 0)$$  where $$b = \frac{2-a - \sqrt{4a - 3a^2}}{2}$$ is the solution to $f(a) = f(b)$ for which $b > a$.  The resulting area of the rectangle is
$$R(a) = a(1-a)^2 \cdot (b-a) = \frac{1}{2}\left( a(1-a)^2 \left(2-3a - \sqrt{4a-3a^2}\right)\right).$$  A tedious differentiation with respect to $a$ yields
$$R'(a) = 1 - 7a + 12a^2 - 6a^3 - (3 - 10a + 6a^2)(1-a) \sqrt{\frac{a}{4-3a}}.$$  So any critical points must satisfy $R'(a) = 0$, or $$(1-7a+12a^2-6a^3)^2(4-3a) = (3-10a+6a^2)^2(1-a)^2 a.$$  This eventually leads to the condition
$$0 = (1-a)^3 (1 - 14a + 58a^2 - 84a^3 + 36a^4).$$  Since $a < 1$, this latter factor yields the minimal polynomial $f(z) = 1 - 14z + 58z^2 - 84z^3 + 36z^4$ for $a$, which actually admits a solution by square roots.  We find this by presuming that $f$ admits a factorization into quadratics of the form
$$f(z) = (1 + pz + 6z^2)(1 + qz + 6z^2)$$
for undetermined constants $p, q$.  Then expanding and equating like coefficients yields the conditions $$p+q = -14, \quad pq + 12 = 58.$$  This gives $(p,q) = -7 \pm \sqrt{3}$ in any order, hence the roots of $f$ are
$$z \in \left\{ \frac{-p \pm \sqrt{p^2 - 24}}{12}, \quad \frac{-q \pm \sqrt{q^2 - 24}}{12} \right\}.$$
Numeric evaluation of the four roots shows that only one satisfies $0 < a < b < 1$, namely $$a = \frac{7 + \sqrt{3} - \sqrt{14(2+\sqrt{3})}}{12} = \frac{7 + \sqrt{3} - \sqrt{7} - \sqrt{21}}{12} \approx 0.12531031679570389137.$$
The corresponding maximum area is then $$R(a) = \frac{7 \sqrt{7} - 10}{108 \sqrt{3}} \approx 0.045547906756179342935.$$
By no means do I claim that this is the most elegant or simplest solution; it is simply what I was able to compute on a first try at this question.
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