Maximum Sphere

For an n-sphere , its surface area is given by:

$\large \displaystyle S_n(R) = \frac{ 2 \pi ^{\frac{n}{2}} }{ \Gamma(\frac{n}{2}) } R^n$

Where $\Gamma(n)$ is the Gamma function . Specifically, for an unit n-sphere $(R=1)$ and since in this problem we're looking for an (n+1)-sphere:

$\large \displaystyle S(n+1) = \frac{ 2 \pi ^{\frac{n+1}{2}} }{ \Gamma(\frac{n+1}{2}) }$

Differentiating it with respect to $n$ and making it equal to zero:

$\large \displaystyle \frac{dS(n+1)}{dn} = \frac{2 \pi ^{\frac{n+1}{2}} \cdot \ln(\pi) \cdot \frac12 \cdot \Gamma(\frac{n+1}{2}) - 2 \pi ^{\frac{n+1}{2}} \cdot \Gamma(\frac{n+1}{2}) \cdot \psi (\frac{n+1}{2}) \cdot \frac12}{\left [ \Gamma(\frac{n+1}{2}) \right ] ^2 } = 0$

$\large \displaystyle \psi \left ( \frac{n+1}{2} \right )= \ln(\pi)$

Where $\psi(n)$ is the Digamma function . Solving numerically for positive $n$ , this gives:

$\large \displaystyle n+1 \approx 7.2569$

Since $n+1$ is an integer, it should be $7$ or $8$ . Calculating both:

$\large \displaystyle S(7) = \frac{16 \pi^3}{15} \approx 33.0734$

$\large \displaystyle S(8) = \frac{\pi^4}{3} \approx 32.4697$

So, the maximum occurs at $\large \displaystyle n + 1= 7$ , or $\color{#3D99F6} \boxed{\large \displaystyle n = 6}$