# 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.

• It certainly requires derivatives. And several more advanced tools, actually. See here.
– user228113
Apr 6, 2017 at 19:32
• You could start by appealing to intuition, that a concave volume doesn't meet the criteria. Any concavities reduce the volume but don't reduce the surface area. So you are left with proving the assertion for a convex volume.
– Χpẘ
Apr 6, 2017 at 20:38

I recommend reading the first part of The Brunn-Minkowski inequality for nilpotent groups by Terence Tao. He first considers the simple one-dimensional Brunn-Minkowski inequality, then proves the Prékopa-Leindler inequality, which again implies the full version of Brunn-Minkowski:

Brunn-Minkowski inequality. Let $A,B$ be two nonempty bounded open subsets of $\mathbb{R}^d$. Denote the Minkowski sum $$A+B = \{ a+b : a \in A, \ b \in B \}$$ and let $|\cdot|$ be the $n$-dimensional Lebesgue measure. Then $$|A+B|^n \ge |A|^{1/n} + |B|^{1/n}.$$

The isoperimetric inequality follows. If $A \subseteq \mathbb{R}^d$ has the same volume $\omega_n$ as the unit ball and $B = B(0,r)$, then $A+B$ equals $A_r$ - the $r$-neighborhood of $A$. Thus Brunn-Minkowski reads $$|A_r|^{1/n} \ge |\omega_n|^{1/n} + |\omega_n r^n|^{1/n} = \omega_n^{1/n} (1+r),$$ then by taking the $n$-th power and substracting $|A|$, $$\frac{|A_r|-|A|}{r} \ge \frac{\omega_n ((1+r)^n - 1)}{r}.$$ If we now take $r \to 0$, the left-hand side converges to the area of $\partial A$ if only $A$ is sufficiently regular (or if we take this as the definition of area), while the right-hand side converges to $n \omega_n$, which is the area of the unit ball. Hence $$\mathrm{Area}(\partial A) \ge \mathrm{Area}(\partial B(0,1)).$$

Of course this doesn't show that the sphere is the unique minimizer. The discussion linked by G. Sassatelli deals with this issue.

• Actually, this approach is even more elementary, as it doesn't involve any derivatives. Apr 6, 2017 at 22:08

The following isn't a proof, but a demonstration a high school student could understand.

First, based on intuition, demonstrate that a volume that minimizes surface area is convex. (Convex in this case meaning all straight lines between any two points in the volume are wholly contained within the volume). If there were any concavities they would reduce the volume, but not reduce the surface area.

Second, imagine the volume in 3D Cartesian space, with the origin ($x=y=z=0$) contained within the volume.

Further imagine bisecting the volume with the $xy$ plane, such that the bisected parts have equal volume. To make the two bisected parts have equal volume, the volume can be translated along the $z$ axis. The upper half volume's points all have non-negative $z$ coordinates.

Now define a function $f(x,y)$ whose value is the distance from the origin to the surface of the volume directly above that $x,y$.

Now imagine integrating $f(x,y)$ over the intersection of the volume and the $xy$ plane, to get the volume of the upper half of the volume. Describe the integration using the concept of the limit of the summation of slivers of volumes. Each sliver could be like a square pyramid with the apex at the origin, each side an (very thin) isoceles triangle, and the base a square. Consider just a single sliver whose base square is centered around the $z$ axis (the pyramid will be upside down).

For this sliver you want to maximize the ratio of the volume to the area of the square. Using $dV$ to represent the volume of the sliver and $dA$ to represent the area of the square at the top of the sliver and $h$ to represent the length of the side of the square:$$dA = h^2$$$$dV = 1/3*dA*f(0,0)$$$$\frac{dV}{dA} = 1/3*f(0,0)$$

So the ratio of the volume of this sliver to the surface area is proportional to the value of $f(x,y)$ at the origin. Based on this alone, it doesn't matter what the function $f(x,y)$ is and the volume could be any convex shape.

However, you want the $\frac{dV}{dA}$ ratio to be maximized no matter which way the sliver of volume points. A sphere achieves this. Imagine that you have two adjacent slivers. If one sliver is longer than the other then you can imagine a sliver between these two whose base is "slanted", the lower end being towards the short sliver on one side and the higher end being towards the sliver on the other side. This slanted base has a bigger area then its neighbors' bases (assuming they are not also slanted). While its volume is larger than its shorter neighbor the ratio between its volume and its surface area is smaller.

This argument, I suspect, is kind of an intuitional equivalent to the calculus of variations argument that the first comment to the OP points to. BTW, if you look at the answer that comment points to, you'll see it stated that this is not an easy problem.

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:

1. Now that, the statement involves two variables area and perimeter, we have to formulate an equation that involves these two.
2. 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:

1. $$y_1 = \sin x \times \cos x$$
2. $$y_2 = x$$

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:

1. 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.

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