Shape Complex

Of all shapes, the total surface area = lateral surface area + 2(base area).

The lateral surface = (height) $\times$ (base perimeter)

Since the height and the base perimeter given in the question are equal, all of those shapes have the same lateral surface area.

Now for these regular polygonal and circular bases, we can evaluate the base areas with perimeter $P$ by dividing them into triangle units with common centroids (( $H_{n}$ denotes the blue height of the triangle subunit) :

For example, the area of the triangular base = 3 $\times$ (area of the triangle unit) = $3\times \dfrac{1}{2} \times \dfrac{P}{3}\times H_{1}$ = $\dfrac{P}{2} \times H_{1}$

Similarly, for the square base, its area = 4 $\times$ (area of the triangle unit) = $4\times \dfrac{1}{2} \times \dfrac{P}{4}\times H_{2}$ = $\dfrac{P}{2} \times H_{2}$

And we can yield similar results for the polygonal bases and the circle, and that demonstrates that the base area varies directly with the internal height $H_{n}$ : the higher the height is, the more the area gets.

As we can see, the blue internal height increases as the number of sides of the polygon increases. In fact, as the number of sides grows, the internal height becomes closer in length with line from the centroid to the vertex, and they will become equal once the shape reaches infinite sides or becomes a circle itself.

Thereby, the circle has its radius equal to such internal height and so will yield the most area, given the same perimeter compared to other shapes.

As a result, the cylinder will have the most total surface area.

Since the solids have the same height and base perimeter, the surface area of the sides are the same. The differences in surface area depend on the base area. The area of an $n$ -sided regular polygon is given by:

$\begin{aligned} A_n & = \frac 12 n\color{#3D99F6}{r_n}^2 \sin \frac {2\pi}n & \small \color{#3D99F6}{\text{where }r_n \text{ is the circumradius of the polygon.}} \\ & = \frac n2 \left(\color{#3D99F6}{\frac p{2n \sin \frac \pi n}}\right)^2 \sin \frac {2\pi}n & \small \color{#3D99F6}{\text{where }p \text{ is the perimeter.}} \\ & = \frac {p^2}{4n} \cot \frac \pi n \\ \implies A_n & = \frac {p^2}{4n \tan \frac \pi n} \end{aligned}$

Consider

$\begin{aligned} n \color{#3D99F6}{\tan \frac \pi n} & = n \color{#3D99F6}{\left(\frac \pi n + \frac {\pi^3}{3n^3} + \frac {2\pi^5}{15n^5} + \cdots \right)} & \small \color{#3D99F6}{\text{By Maclaurin series}} \\ & = \pi + \frac {\pi^3}{3n^2} + \frac {2\pi^5}{15n^4} + \cdots \end{aligned}$

This implies that as $n$ increases, $n\tan \frac \pi n$ decreases and $A_n$ increases. Therefore, $A_n$ and the overall surface area is largest for $\boxed{\text{Cylinder}}$ as $n \to \infty$ for a circle.

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Just to check the result of $A_n$ for a circle.

$\begin{aligned} A_\infty & = \lim_{n \to \infty} \frac {\color{#3D99F6}{p}^2}{4n \tan \frac \pi n} & \small \color{#3D99F6}{\text{Let }p = 2\pi r} \\ & = \frac {4\pi^2 r^2}{4\color{#3D99F6}{\pi}} & \small \color{#3D99F6}{\text{Note that }\lim_{n \to \infty} n\tan \frac \pi n = \pi} \\ & = \pi r^2 \end{aligned}$