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)


\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).

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





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.