7.11. Extremal ProblemsApplying math often comprises the task to find a certain setting for some parameters of a process to ensure an optimal result (e.g. shortest distance, highest profit, lowest consumption). In many cases our techniques of detecting global extreme points will enable us to find a calculative solution. For a simple example to start with we consider rectangles with a fixed perimeter U. There are lots of those rectangles and they all differ in their area as can be checked by dragging the blue anchor in the sketch below.
To that end we survey all the areas encountered, in other words we consider the function A given by
$A(x,y)=x\cdot y$.[1]
The variables x and y however are not independent from each other as the side condition, that is setting the perimeter $U>0$ as fixed, forces x and y into a distinct relationship:
$2x+2y=U\text{\hspace{1em}}\iff \text{\hspace{1em}}y=\frac{U}{2}x$.[2]
Thus we may substitute the variable y in [1] by some term of x. Considering that the width x may only take values between 0 and $\frac{U}{2}$ we get the following definition for the target function $A:[\mathrm{0,}\frac{U}{2}]\to \mathbb{R}$
$A(x)=x\cdot (\frac{U}{2}x)={x}^{2}+\frac{U}{2}x$.
Looking for a greatest area now means searching for a global maximum for A. Such a maximum point will exist as A is continuous on a closed interval (see[6.6.5]). It could be one of the boundary points or a local maximum point in the interior of the domain of A. The latter case is easily dealt with using our criteria for twice differentiable functions: As
$A}^{\prime}(x)=0\text{\hspace{1em}}\iff \text{\hspace{1em}}2x+\frac{U}{2}=0\text{\hspace{1em}}\iff \text{\hspace{1em}}x=\frac{U}{4$
and ${A}^{\prime}}^{\prime$ we find a single local maximum in the interior namely at $\frac{U}{4}$ with a value of
$A(\frac{U}{4})=\frac{{U}^{2}}{16}>0$.
By comparison with the boundary values $A(0)=0$ and $A(\frac{U}{2})=0$ we finally see that the rectangle with $x=\frac{U}{4}$ and $y=\frac{U}{4}$ (according to [2]) has the greatest area. Thus we proved the well known fact:
This result is a special case of the so called isoperimetric problem, i.e. to find among all laminae with a fixed (appropriate) perimeter the one(s) with the greatest area. Although it is quite obvious that this will be a circle it is far from simple to prove this. For more information see Viktor Blasjö: The Isoperimetric Problem, Amer. Math. Monthly 112, pp. 526566. A similar problem, often encountered in economic activities, is called the packaging problem: How to gain a maximal volume from a fixed surface? As an example we create a box (no lid) of greatest volume from a rectangular cardboard sized a×b ($a\ge b>0$). We just cut out four squares of width x as shown in the drawing and then flip up the resulting straps. Our cardboard is sized a = 240px and b = 180px.
introduces the target function $V:[\mathrm{0,}\frac{b}{2}]\to \mathbb{R}$ directly without quoting the
side condition
i A detailed notation, like for instance with side condition "length = a − 2x and width = b − 2x", turns out to be unnecessary laborious in straightforward situations like this.
$\begin{array}{l}V(x)=4{x}^{3}2(a+b){x}^{2}+abx\\ {V}^{\prime}(x)=12{x}^{2}4(a+b)x+ab\\ {{V}^{\prime}}^{\prime}(x)=24x4(a+b)\end{array}$
we get: ${V}^{\prime}(x)=0\text{\hspace{1em}}\iff \text{\hspace{1em}}x=\frac{1}{6}(a+b)\pm \frac{1}{6}\sqrt{{(a+b)}^{2}3ab}=\frac{1}{6}(a+b\pm \sqrt{{(ab)}^{2}+ab})$. As $\frac{1}{6}(a+b+\sqrt{{(ab)}^{2}+ab})\ge \frac{1}{6}(a+b+\sqrt{ab})\ge \frac{1}{6}(b+b+\sqrt{bb})=\frac{b}{2}$,
the first root does not belong to $]0,\frac{b}{2}[$ in contrast to the second one because here we have:
$\begin{array}{ll}0=\frac{1}{6}(a+b\sqrt{{(a+b)}^{2}})\hfill & <\frac{1}{6}(a+b\sqrt{{(a+b)}^{2}3ab})\hfill \\ \hfill & =\frac{1}{6}(a+b\sqrt{{(ab)}^{2}+ab})\hfill \\ \hfill & <\frac{1}{6}(a+b\sqrt{{(ab)}^{2}})\hfill \\ \hfill & =\frac{b}{3}<\frac{b}{2}\text{.}\hfill \end{array}$[4]
Furthermore ${{V}^{\prime}}^{\prime}(\frac{1}{6}(a+b\sqrt{{(ab)}^{2}+ab}))=\sqrt{{(ab)}^{2}+ab}<0$ in fact proves the second root to be a local maximum point. And according to [4] twice of its value is less than b, and therefor less than a as well. Thus [3] guarantees
$V(\frac{1}{6}(a+b\sqrt{{(ab)}^{2}+ab}))>0$,
which means that $\frac{1}{6}(a+b\sqrt{{(ab)}^{2}+ab})$ leaves the comparison test with both boundary values $V(0)=0=V(\frac{b}{2})$ as a global maximum point. In our example this calculates to a (rounded) value of 34px.
In physics the behaviour of systems is often controlled in such a way that a certain variable reaches its minimum value. Forced by energy balancing the shape of a soap bubble for instance will always have a minimal surface and thus will come as a ball. Often this minimization principle is also found as the general idea behind the laws of physics. We will demonstrate this with the law of reflection: If a signal is reflected off a mirror plane the angle of incidence and angle of reflexion are equal. We will now prove that this holds for a reflection if and only if the signal follows the shortest way from sender to receiver.
