September 2020

Shown below is a cylinder shaped wine barrel.

Fig. 1

From Fig. 1, we see that

$d^2 = (\frac{h}{2})^2 + (2r)^2\implies r^2=\frac{d^2}{4}-\frac{h^2}{16}\quad\quad\quad(1)$

and so,

$V=\pi r^2 h \overset{(1)}{\implies} V=\pi (\frac{d^2h}{4} - \frac{h^3}{16})\quad\quad\quad(2)$

Kepler’s “Wine Barrel Problem” can be stated as:

If $d$ is fixed, what value of $h$ gives the largest volume of $V$ ?

Kepler conducted extensive numerical studies on this problem. However, it was solved analytically only after the invention of calculus.

In the spring of 2012, while carrying out a research on solving maximization/minimization problems, I discovered the following theorem:

Theorem-1. For positive quantities $a_1, a_2, ..., a_n, c_1, c_2, ..., c_n$ and positive rational quantities $p_1, p_2, ..., p_n$ , if $c_1a_1+c_2a_2+...+ c_na_n$ is a constant, then $a_1^{p_1}a_2^{p_2}...a_n^{p_n}$ attains its maximum if $\frac{c_1a_1}{p_1} = \frac{c_2a_2}{p_2} = ... = \frac{c_na_n}{p_n}$ .

By applying Theorem-1, the “Wine Barrel Problem” can be solved analytically without calculus at all. It is as follows:

Rewrite (2) as

$V = \sqrt{16} \pi (\frac{h^2}{16})^{\frac{1}{2}}(\frac{d^2}{4} - \frac{h^2}{16})^1.\quad\quad\quad(3)$

Since

$\frac{h^2}{16} >0, \frac{d^2}{4} - \frac{h^2}{16} >0$ and $1\cdot \frac{h^2}{16} + 1\cdot(\frac{d^2}{4} -\frac{h^2}{16}) = \frac{d^2}{4}$ , a constant,

by Theorem-1, when

$\frac{1\cdot\frac{h^2}{16}}{\frac{1}{2}} = \frac{1\cdot(\frac{d^2}{4}-\frac{h^2}{16})}{1},\quad\quad\quad$

$\frac{\frac{h^2}{16}}{\frac{1}{2}} = \frac{d^2}{4}-\frac{h^2}{16},\quad\quad\quad(4)$

$V$ (see (3)) attains its maximum.

Solving (4) for positive $h$ , we have

$h = \frac{2}{\sqrt{3}}d.$

Discovered from the same research is another theorem for solving minimization problem without calculus:

Theorem-2. For positive quantities $a_1, a_2, ..., a_n, c_1, c_2, ..., c_n$ and positive rational quantities $p_1, p_2, ..., p_n$ , if $a_1^{p_1}a_2^{p_2}...a_k^{p_k}$ is a constant, then $c_1a_1+c_2a_2+...+c_na_n$ attains its minimum if $\frac{c_1a_1}{p_1} = \frac{c_2a_2}{p_2} = ... = \frac{c_na_n}{p_n}$ .

Let’s look at an example:

Problem: Find the minimum value of $\frac{1}{\sqrt[3]{x}} + 27x$ for $x>0$ .

Since for $x >0, \frac{1}{\sqrt[3]{x}} >0, 27x >0$ and $(\frac{1}{\sqrt[3]{x}})^3\cdot 27x = 27$ , a constant,

by Theorem-2, when

$\frac{\frac{1}{\sqrt[3]{x}}}{3} = 27x,\quad\quad\quad(5)$

$\frac{1}{\sqrt[3]{x}} + 27x$ attains its minimum.

Solving (5) for $x$ yields

$x = \frac{1}{27}$ .

Therefore, at $x = \frac{1}{27}\approx 0.03703, \frac{1}{\sqrt[3]{x}} + 27x$ attains its minimum value $\sqrt[3]{27}+27\cdot\frac{1}{27}= 3+1=4$ (see Fig. 2).