Shown below is a cylinder shaped wine barrel.
Fig. 1
From Fig. 1, we see that

and so,

Kepler’s “Wine Barrel Problem” can be stated as:
If
is fixed, what value of
gives the largest volume of
?
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
and positive rational quantities
, if
is a constant, then
attains its maximum if
.
By applying Theorem-1, the “Wine Barrel Problem” can be solved analytically without calculus at all. It is as follows:
Rewrite (2) as

Since
and
, a constant,
by Theorem-1, when
or

(see (3)) attains its maximum.
Solving (4) for positive
, we have

Discovered from the same research is another theorem for solving minimization problem without calculus:
Theorem-2. For positive quantities
and positive rational quantities
, if
is a constant, then
attains its minimum if
.
Let’s look at an example:
Problem: Find the minimum value of
for
.
Since for
and
, a constant,
by Theorem-2, when
![\frac{\frac{1}{\sqrt[3]{x}}}{3} = 27x,\quad\quad\quad(5) \frac{\frac{1}{\sqrt[3]{x}}}{3} = 27x,\quad\quad\quad(5)](https://s0.wp.com/latex.php?latex=%5Cfrac%7B%5Cfrac%7B1%7D%7B%5Csqrt%5B3%5D%7Bx%7D%7D%7D%7B3%7D+%3D+27x%2C%5Cquad%5Cquad%5Cquad%285%29&bg=ffffff&fg=444444&s=0&c=20201002)
attains its minimum.
Solving (5) for
yields
.
Therefore, at
attains its minimum value
(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
for
.
Theorem-2 is not applicable here (see Exercise-1). To solve this problem, we proceed as follows:
From
, we have

and so,
.
That is,
.
Or,




i.e.
.
Since
,
,
with the “=” sign in “
” holds at
.
Therefore,
attains its minimum -54 at
(see FIg. 3).
Fig. 3
Exercise-1 Explain why Theorem-2 is not applicable to finding the minimum of
for
.