cuboid with least surface
CuboidWithLeastSurface
2007-07-15 17:42:51
2007-09-04 10:36:33
Example
extremum points of function of several variables
cuboid
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\newtheorem*{thmplain}{Theorem}
Let us determine among all {\em cuboids} (i.e. rectangular parallelepipeds) with a given volume $k^3$ such one which has the least surface area.
Let the three edges of the cuboid beginning from a vertex be $x$, $y$ and $z$; then we must start from the condition\, $xyz = k^3$,\, whence\, $z = \frac{k^3}{xy}$. We get the expression
\begin{align}
f(x,\,y) := 2(yz\!+\!zx\!+\!xy) = 2\!\left(\!xy+\frac{k^3}{x}+\frac{k^3}{y}\!\right)
\end{align}
for the whole area of the \PMlinkescapetext{surface} of the cuboid. Thus we have to make\, $f(x,\,y)$ a minimum, when only the positive values of $x$ and $y$ can be taken into consideration.
The function $f$ and its first order partial derivatives are continuous for all positive $x$ and $y$. According to the \PMlinkescapetext{theorem} of the \PMlinkname{parent entry}{ExtremumPointsOfFunctionOfSeveralVariables}, a minimum can occur only when simultaneously
\begin{align*}
\begin{cases}
{f'_x(x,\,y) = y-\frac{k^3}{x^2} = 0},\\
{f'_y(x,\,y) = x-\frac{k^3}{y^2} = 0}.
\end{cases}
\end{align*}
These equations are true only for\, $x = y = k$,\, i.e. for the case that the cuboid is a cube.
We can infer that a cube has the least area. In fact, we see from (1) that\, $f(x,\,y)\to\infty$\, as\, $x\to 0$\, or $y\to 0$\, or\, $xy\to\infty$;\, therefore there exist a small positive number $m$ and a big positive number $M$ such that outside and on the boundary of the region resembling a triangle and bounded by the lines \, $x = m$\, and\, $y = m$\, and the rectangular hyperbola \, $xy = M$,\, the value of\, $f(x,\,y)$\, is always greater than in the point \, $(k,\,k)$\, inside this region. Thus the function gets its smallest value in an interior point of the region, and this point must be\, $(k,\,k)$\, since it is the only point where $f'_x$ and $f'_y$ both vanish.\\
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