Solving Kepler’s “Wine Barrel Problem” without Calculus

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

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

or

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

$V$ (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).

Fig. 2

Nonetheless, neither Theorem-1 nor Theorem-2 is a silver bullet for solving max/min problems without calculus. For example,

Problem: Find the minimum value of $x^3-27x$ for $x>0$.

Theorem-2 is not applicable here (see Exercise-1). To solve this problem, we proceed as follows:

From $(x-3)^3 = x^3-9x^2+27x-27$, we have

$(x-3)^3+9x^2+27 = x^3+27x$

and so,

$(x-3)^3+9x^2+27-54x =x^3-27x$.

That is,

$(x-3)^3+9(x^2-6x+3)=x^3-27x$.

Or,

$x^3-27x= (x-3)^3+9(x^2-6x+9-6)$

$=(x-3)^3+9((x-3)^2-6)=(x-3)^3+9(x-3)^2-54$

$=-54+(x-3)^3+9(x-3)^2=-54+(x-3)^2(x-3+9)$

$=-54+(x-3)^2(x+6)$

i.e.

$x^3-27x = -54 +(x-3)^2(x+6)$.

Since $x>0, x+6>0, (x-3)^2 \ge 0 \implies (x-3)^2(x+6) \ge 0$,

$x^3-27x = -54 + (x-3)^3(x+6) \ge -54$,

with the “=” sign in “$\ge$” holds at $x=3$.

Therefore, $x^3-27x$ attains its minimum -54 at $x=3$ (see FIg. 3).

Fig. 3

Exercise-1 Explain why Theorem-2 is not applicable to finding the minimum of $x^3-27x$ for $x>0$.

“Mine’s Bigger!”

Question: Which one is bigger, $e^\pi$ or $\pi^e?$

Consider function

$f(x) = \frac{\log(x)}{x}$.

We have

$f'(x) = \frac{1-\log(x)}{x^2} \implies f'(e)=\frac{1-\log(e)}{e^2} \overset{\log(e)=1}{=} 0\quad\quad\quad(1)$

and

$f''(x) = \frac{2\log(x)}{x^3}-\frac{3}{x^3}\implies f''(e) = \frac{2\log(e)}{e^3}-\frac{3}{e^3}= -\frac{1}{e^3}<0.\quad\quad\quad(2)$

From (1) and (2), we see that

$\frac{\log(x)}{x}$ attains its global maximum at $x=e$

which means

$\forall x >0, x \neq e \implies \frac{\log(x)}{x} < \frac{\log(e)}{e}$.

When $x=\pi$,

$\frac{\log(\pi)}{\pi} < \frac{\log(e)}{e}$

or,

$e\log(\pi) < \pi\log(e)$.

It follows that

$\log(\pi^e) < \log(e^\pi).\quad\quad\quad(3)$

Since $\log(x)$ is a monotonic increasing function (see Exercise-1), we deduce from (3) that

$e^\pi > \pi^e$

Exercise-1 Show that $\log(x)$ is a monotonic increasing function.