Maximise it!

Using the same notations as David, we want to maximize twice the area of the trapezoid $ABDE$, which is equivalent of maximizing the difference between the area of the triangle $ACE$ and the area of the triangle $BCD.$ Let $T$ denote this difference.

Since $CE = 52 \div2= 26 ,$ let $y = CD ,$ then $DE = 26 - y .$

Because the triangles $ACE$ and $BCD$ are similar, then $\dfrac yx = \dfrac{26}{12 + x} \quad \implies\quad y = \dfrac{26x}{x + 12} .$

Let $\theta = \measuredangle BCD ,$ then $\cos \theta = x \div \dfrac{26x}{x+12} = \dfrac{x+12}{26} .$

$\begin{array}{ r c l } T &=& \dfrac 12 (AC)(CE) \sin \theta - \dfrac12 (BC)(CD)\sin\theta \\ \phantom 0 \\ &=& \dfrac12 (x+12) \cdot 26 \sin \theta - \dfrac12 \cdot \dfrac{26x}{x+12} \cdot x \cdot \sin \theta \\ \phantom 0 \\ &=& \dfrac{26}2 \sin \theta \left [ (x+12) - \dfrac{x^2}{x+12} \right ] \\ \phantom 0 \\ &=& 13 \sqrt{1 - \cos^2 \theta} \cdot \dfrac1{x+12} \cdot [ (x+12)^2 - x^2] \\ \phantom 0 \\ \dfrac T{13} &=& \sqrt{1 - \left( \dfrac{x+12}{26} \right)^2 } \cdot \dfrac1{x+12} \cdot (24x + 144) \\ \phantom 0 \\ \dfrac T{13 \cdot 24} &=& \sqrt{1 - \left( \dfrac{x+12}{26} \right)^2 } \cdot \dfrac1{x+12} \cdot (x+ 6) \\ \phantom 0 \\ &=& \sqrt{1 - \left( \dfrac{x+12}{26} \right)^2 } \cdot \dfrac1{x+12} \cdot [ (x+12) - 6 ] \end{array}$

For simplicity sake, let $y = x + 12$,

Equivalently, we want to maximize the expression

$\left( \dfrac T{13\cdot 24} \right)^2 = \left[ 1 - \left(\dfrac y{26}\right)^2 \right ] \cdot \dfrac1{y^2} \cdot (y - 6)^2$

Multiplying both sides by $26^2 ,$ the expression we want to maximize is

$f(y) := \dfrac{26^2 - y^2}{y^2} (y - 6)^2 = \left(\dfrac{26^2}{y^2} - 1 \right) (y -6)^2$

A simple chain rule shows that $f'(y) = -\dfrac{2(y-6)(y^3 - 4056)}{y^2} .$

At the critical point, $f'(y) = 0 \implies y = 6, 4056^{1/3} .$ But $f(6) < f \left( 4056^{1/3} \right) ,$ so the expression is maximized at $y = 4056^{1/3} .$

Equivalently, the area on the pentagon is maximized when $x = y - 12 =\boxed{ \sqrt[3]{4056} - 12 } .$

Label the diagram as follows, and let $\theta = \angle ECA$:

By symmetry, $EC = \frac{1}{2} \cdot 52 = 26$ and $AB = \frac{1}{2} \cdot 24 = 12$.

From $\triangle ACE$, $AC = 26 \cos \theta = x + 12$, so $x = 26 \cos \theta - 12 = BC$.

Since $\triangle ACE \sim \triangle BCD$ by AA similarity, $CD = \cfrac{EC \cdot BC}{AC} = \cfrac{26 \cos \theta - 12}{\cos \theta}$.

The area of the pentagon $A_{\text{pent}}$ is then:

$= 2 \cdot A_{ABDE}$

$= 2(A_{\triangle ACE} - A_{\triangle BCD})$

$= 2(\frac{1}{2} \cdot AC \cdot CE \cdot \sin \theta - \frac{1}{2} \cdot BC \cdot CD \cdot \sin \theta)$

$= 2(\frac{1}{2} \cdot 26 \cos \theta \cdot 26 \cdot \sin \theta - \frac{1}{2} \cdot (26 \cos \theta - 12) \cdot \cfrac{26 \cos \theta - 12}{\cos \theta} \cdot \sin \theta)$

$= 26^2 \cos^2 \theta \tan \theta - (26 \cos \theta - 12)^2 \tan \theta$

$= (624 \cos \theta - 144) \tan \theta$

$= 48(13 \cos \theta - 3) \tan \theta$

$= 48(13 \sin \theta - 3 \tan \theta)$

And its derivative $\cfrac{dA_{\text{pent}}}{d\theta}$ is then:

$= 48(13 \cos \theta - 3 \sec^2 \theta)$

$= 48(13 \cos^3 \theta \sec^2 \theta - 3 \sec^2 \theta)$

$= 48 \sec^2 \theta (13 \cos^3 \theta - 3)$

The maximum area occurs when $\cfrac{dA_{\text{pent}}}{d\theta} = 48 \sec^2 \theta (13 \cos^3 \theta - 3) = 0$, which is when $\cos \theta = \sqrt[3]{\cfrac{3}{13}}$.

That makes the value of $x$ for which the area of pentagon is a maximum $x = 26 \cos \theta - 12 = 26 \sqrt[3]{\cfrac{3}{13}} - 12 = 2 \sqrt[3]{507} - 12 \approx \boxed{3.948}$.