Maximum surface area of a variable polyhedron

The surface area $A$ of the polyhedron only depends on the side length of the top square $x$ . It has a minimum of $200$ when $x=0$ . The additional surface area is the square on top $x^2$ and the area of the four slanting triangles which we can find using Heron's formula , where its semi-perimeter $s = \frac {5\sqrt 2+5\sqrt 2+x}2 = 5\sqrt 2 + \frac x2$ . The surface area $A$ is given by:

$\begin{aligned} A & = x^2 + 4\sqrt{\left(5\sqrt 2 + \frac x2\right)\left(5\sqrt 2 - \frac x2\right)\left(\frac x2\right)^2} + 200 \\ & = x^2 + x \sqrt{200-x^2} + 200 & \small \blue{\text{To find }\max(A)} \\ \frac {dA}{dx} & = 2x + \frac {200x - 2x^3}{x\sqrt{200-x^2}} \\ & = \frac {2x\sqrt{200-x^2}+200-2x^2}{\sqrt{200-x^2}} & \small \blue{\text{Putting }\frac {dA}{dx} = 0} \end{aligned}$

$\begin{aligned} 2x\sqrt{200-x^2} & = 2x^2 - 200 \\ x\sqrt{200-x^2} & = x^2 - 100 \\ 200x^2 - x^4 & = x^4 - 200x^2 + 10000 \\ 2x^4 - 400x^2 + 10000 & = 0 \\ x^4 - 200x^2 + 5000 & = 0 \end{aligned}$

$\begin{aligned} \implies x^2 & = \frac {\pm \sqrt{200^2 - 4(5000)}+200}2 = \pm 50\sqrt 2 + 100 \\ & = 100\left(1+\frac 1{\sqrt 2} \right) & \small \blue{100 - 50\sqrt 2 \text{ is too small.}} \\ x & = 10 \sqrt{1+\frac 1{\sqrt 2}} \end{aligned}$

$\begin{aligned} \implies A_{\max} & = 100\left(1+\frac 1{\sqrt 2}\right) + 10 \sqrt{1+\frac 1{\sqrt 2}}\cdot \sqrt{200-50\sqrt 2 - 100} + 200 \\ & = 200 + 50 \sqrt 2 + 10 \sqrt{1+\frac 1{\sqrt 2}}\cdot \sqrt{100-50\sqrt 2} \\ & = 300 + \frac {100}{\sqrt 2} + 10 \sqrt{1+\frac 1{\sqrt 2}}\cdot 10 \sqrt{1-\frac 1{\sqrt 2}} \\ & = 300 + \frac {100}{\sqrt 2} + \frac {100}{\sqrt 2} = 300 + 100\sqrt 2 \\ & \approx \boxed{441.42} \end{aligned}$