A problem on Minima. Proving sphere has minimum surface area for a given volume Has the below question been answered here before?
Prove: For a given enclosed volume, a sphere has minimum surface area.
Please provide link or ways to solve it. I know it is a problem of Minima and involves finding derivative and second derivatives. However, how to to frame the leading equations which has to be differentiated?
I am looking for a way to solve only using High School level Calculus. I think there must be an answer.

To prove: For a given enclosed volume, a sphere has minimum surface area.
It is the same as

The sphere has the minimum surface area for a given volume.
OR
The sphere has the maximum volume for a given surface area.

Let's simplify the three-dimensional problem into two-dimensions by considering the following statement:

The circle has the maximum area for a given perimeter (boundary).

Few things to consider while solving the problem:

*

*Now that, the statement involves two variables area and perimeter, we have to formulate an equation that involves these two.

*As the problem also involves different objects like circle, and let's say a square, we also have to incorporate the shape variable

Let's consider:

*

*$A$ and $P$ represent the area and perimeter of a 2D object in a general.



*

*To be specific, let's say $A_s$ and $P_s$ corresponds to the area and perimeter of a square. $A_c$ and $P_c$ corresponds to the area and
perimeter of a circle.


*$a$ is the length of each side of a square. $r$ is the radius of the circle.
The area of the circle will be $A_c = \pi r^2$ and the perimeter will
be $P_c= 2 \pi r$. Thus, $ A_c = \pi \times \bigg( \frac{P_c}{2 \pi}
  \bigg)^2  \Rightarrow A_c = \frac{P_c^2}{4 \pi} \Rightarrow A_c = k_c
  P_c^2 $
Similarly, for a square the formulation will be $ A_s = \bigg(
  \frac{P_s}{4} \bigg)^2 = k_s P_s^2$
Such formulations doesn't consider the shape of the circle or square.

To incorporate the shape into these equations, we have to consider the number of edges ($n$) the objects have. For instance, the square has $4$ edges whereas the circle has $\infty$ edges.
For simplicity, we consider only regular convex polygons.
Let's consider a regular polygon of $n$ sides. Each of its sides has a length $l$. Thus, the perimeter $P = nl \Rightarrow l = \frac{P}{n}$.
The regular hexagon in the picture is only for illustration. The angle subtended by each side of the polygon at the center is $\frac{2 \pi}{n}$ as shown in the following figure.
Regular polygon:

In the $\Delta OMB$,
$$tan \frac{\pi}{n} = \frac{MB}{OM}
\Rightarrow OM = \frac{l/2}{\tan \frac{\pi}{n}}
\Rightarrow OM = \frac{P}{2n \tan \frac{\pi}{n}}$$
Thus, area of the $\Delta OAB$ is
$$ area(\Delta OAB) = \frac{1}{2} \times OM \times MB 
= \frac{1}{2} \times \frac{P}{2n \tan \frac{\pi}{n}} \times l $$
$$ \Rightarrow area(\Delta OAB) = \frac{P^2}{4 n^2 \tan \frac{\pi}{n}} $$
So, area of the polygon will be
$$A = n \times \frac{P^2}{4 n^2 \tan \frac{\pi}{n}} $$
$$A =  \frac{P^2}{4 n \tan \frac{\pi}{n}} $$
Now, we have an expression that correlates the area and perimeter while incorporating shape (number of edges). So, we have to prove that the circle has the maximum area for a given perimeter ($P=$constant) among all the numbers of edges.

For example, a (regular ~ Equilateral) triangle has three sides, square (regular quadrilateral) has four and so on.


We have to prove: The area of the regular polygon increases as we go on increasing the number of edges. Alternatively, the area is highest when the number of edges is the maximum, i.e $n=\infty$, which is the case for a circle.

The maxima (or minima) can be found by considering
$$\frac{dA}{dn} = 0$$
$$\Rightarrow \frac{d}{dn} \Bigg( \frac{P^2}{4 n \tan \frac{\pi}{n}} \Bigg)= 0$$
$$ \Rightarrow \frac{P^2}{4} \times \bigg( \frac{-1}{ (n \tan \frac{\pi}{n} )^2}\bigg) \times \bigg( \tan \frac{\pi}{n} + n \sec ^2 \frac{\pi}{n} \times \frac{- \pi}{n^2} \bigg) =0$$
where $P$ is constant. The above equation simplifies to
$$\sin \frac{\pi}{n} \times \cos \frac{\pi}{n} = \frac{\pi}{n}$$
Considering, $\frac{\pi}{n} = x$, the above equation can be written as $ \sin x \times \cos x = x$. It is a transcendental equation. They cannot be solved deterministically. Their solutions can be approximated using numerical methods. We split the equation into two parts:

*

*$$ y_1 = \sin x \times \cos x$$

*$$y_2 = x$$
If we look at their plots...
Plot of $y_1=sin(x) cos(x)$ and $y_2=x$ to solve $sin(x)cos(x)=x$:

we can see that the solution is $x=0  \Rightarrow n = \frac{\pi}{x} = \infty$
The number of edges required to maximize the area is $\infty$. So, our regular polygon takes the shape of a circle.

Few things to note:

*

*To show that the circle has the minimum boundary for a given area (A=constant) ,  the equation $A =  \frac{P^2}{4 n \tan \frac{\pi}{n}} $ will remain same. It just needs to be rewritten as $$P =  2 \sqrt {n A \tan \frac{\pi}{n}} $$ and $\frac{dP}{dn}=0$ needs to be solved. It also leads to the same solution.


*The same can be extended to 3D for the case of a sphere by considering solid angles.
