7.11. Extremal Problems

Applying 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.

 ▪ This rectangle has a fixed perimeter of 340px. With a width of x =  and a height of  y =   its area calculates to x·y =  We are now going to find a rectangle among those with greatest area.

To that end we survey all the areas encountered, in other words we consider the function A given by

$A\left(x,y\right)=x\cdot y$.

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:

$2x+2y=U\text{ }⇔\text{ }y=\frac{U}{2}-x$.

Thus we may substitute the variable  y in  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  $A:\left[0,\frac{U}{2}\right]\to ℝ$

$A\left(x\right)=x\cdot \left(\frac{U}{2}-x\right)=-{x}^{2}+\frac{U}{2}x$.

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

${A}^{\prime }\left(x\right)=0\text{ }⇔\text{ }-2x+\frac{U}{2}=0\text{ }⇔\text{ }x=\frac{U}{4}$

and ${{A}^{\prime }}^{\prime }$ we find a single local maximum in the interior namely at $\frac{U}{4}$ with a value of

$A\left(\frac{U}{4}\right)=\frac{{U}^{2}}{16}>0$.

By comparison with the boundary values  $A\left(0\right)=0$ and $A\left(\frac{U}{2}\right)=0$ we finally see that the rectangle with $x=\frac{U}{4}$ and $y=\frac{U}{4}$ (according to ) has the greatest area. Thus we proved the well known fact:

Proposition:

 The square has the greatest area among all rectangles with the same perimeter. [7.11.1]

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. 526-566.

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.

If we clip a square of at each corner the box will be a − 2x =  long, b − 2x =  broad and x =  high. Hence it will be

(a − 2x)·(b − 2xx =

big. Setting now

$V\left(x\right)=\left(a-2x\right)\cdot \left(b-2x\right)\cdot x$

introduces the target function  $V:\left[0,\frac{b}{2}\right]\to ℝ$ directly without quoting the side condition i A detailed notation, like for instance V(x) = length·width·x, with side condition "length = a − 2x and width = b − 2x", turns out to be unnecessary laborious in straightforward situations like this.
explicitly. Next we find the local extreme points for V within the interior of the domain interval. From

$\begin{array}{l}V\left(x\right)=4{x}^{3}-2\left(a+b\right){x}^{2}+abx\\ {V}^{\prime }\left(x\right)=12{x}^{2}-4\left(a+b\right)x+ab\\ {{V}^{\prime }}^{\prime }\left(x\right)=24x-4\left(a+b\right)\end{array}$

we get:  ${V}^{\prime }\left(x\right)=0\text{ }⇔\text{ }x=\frac{1}{6}\left(a+b\right)±\frac{1}{6}\sqrt{{\left(a+b\right)}^{2}-3ab}=\frac{1}{6}\left(a+b±\sqrt{{\left(a-b\right)}^{2}+ab}\right)$. As

$\frac{1}{6}\left(a+b+\sqrt{{\left(a-b\right)}^{2}+ab}\right)\ge \frac{1}{6}\left(a+b+\sqrt{ab}\right)\ge \frac{1}{6}\left(b+b+\sqrt{bb}\right)=\frac{b}{2}$,

the first root does not belong to $\right]0,\frac{b}{2}\left[$ in contrast to the second one because here we have:

$\begin{array}{ll}0=\frac{1}{6}\left(a+b-\sqrt{{\left(a+b\right)}^{2}}\right)\hfill & <\frac{1}{6}\left(a+b-\sqrt{{\left(a+b\right)}^{2}-3ab}\right)\hfill \\ \hfill & =\frac{1}{6}\left(a+b-\sqrt{{\left(a-b\right)}^{2}+ab}\right)\hfill \\ \hfill & <\frac{1}{6}\left(a+b-\sqrt{{\left(a-b\right)}^{2}}\right)\hfill \\ \hfill & =\frac{b}{3}<\frac{b}{2}\text{.}\hfill \end{array}$

Furthermore  ${{V}^{\prime }}^{\prime }\left(\frac{1}{6}\left(a+b-\sqrt{{\left(a-b\right)}^{2}+ab}\right)\right)=-\sqrt{{\left(a-b\right)}^{2}+ab}<0$ in fact proves the second root to be a local maximum point.

And according to  twice of its value is less than b, and therefor less than a as well. Thus  guarantees

$V\left(\frac{1}{6}\left(a+b-\sqrt{{\left(a-b\right)}^{2}+ab}\right)\right)>0$,

which means that $\frac{1}{6}\left(a+b-\sqrt{{\left(a-b\right)}^{2}+ab}\right)$ leaves the comparison test with both boundary values $V\left(0\right)=0=V\left(\frac{b}{2}\right)$ 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.

Proposition:  Let α be the angle of incidence and β the angle of reflection of an arbitrary reflection. Then the following holds:

 $\alpha =\beta \text{ }⇔\text{ }$the path of the signal has a minimal length. [7.11.2]

Proof:  We introduce a suitable coordinate plane that locates the sender at (0,s), the receiver at (a,b) and the reflection point at (x,0). For $a,b,s,x>0$ we thus have the following scene : animation ↔ trigonometry s b xa

The length $l\left(x\right)$ of the path depends on the position of the reflection point and is calculated according to Pythagoras' theorem (choose option trigonometry) like this:

$l\left(x\right)=\sqrt{{x}^{2}+{s}^{2}}+\sqrt{{\left(a-x\right)}^{2}+{b}^{2}},\text{ }x\in \left[0,a\right]$

l is twice differentiable and its derivatives are

$\begin{array}{ll}\hfill {l}^{\prime }\left(x\right)=& \frac{x}{\sqrt{{x}^{2}+{s}^{2}}}-\frac{a-x}{\sqrt{{\left(a-x\right)}^{2}+{b}^{2}}}\phantom{\rule{50px}{0ex}}\text{}\hfill \\ \hfill {{l}^{\prime }}^{\prime }\left(x\right)=& \frac{{s}^{2}}{{\sqrt{{x}^{2}+{s}^{2}}}^{3}}+\frac{{b}^{2}}{{\sqrt{{\left(a-x\right)}^{2}+{b}^{2}}}^{3}}>0\phantom{\rule{50px}{0ex}}\text{}\hfill \end{array}$

 now yields the following equivalence for all $x\in \right]0,a\left[$:

$\begin{array}{ll}{l}^{\prime }\left(x\right)=0\text{ }\hfill & ⇔\text{ }\frac{x}{\sqrt{{x}^{2}+{s}^{2}}}=\frac{a-x}{\sqrt{{\left(a-x\right)}^{2}+{b}^{2}}}\hfill \\ \hfill & ⇔\text{ }\mathrm{sin}\alpha =\mathrm{sin}\beta \hfill \\ \hfill & ⇔\text{ }\alpha =\beta \phantom{\rule{50px}{0ex}}\text{}\hfill \\ \hfill & ⇔\text{ }\mathrm{tan}\alpha =\mathrm{tan}\beta \hfill \\ \hfill & ⇔\text{ }\frac{s}{x}=\frac{b}{a-x}\hfill \\ \hfill & ⇔\text{ }x=\frac{a\cdot s}{b+s}\hfill \end{array}$

Due to  we thus know: l has a single local minimum point within $\right]0,a\left[$ namely at $\frac{a\cdot s}{b+s}$ i Consider:  $0<\frac{a\cdot s}{b+s}<\frac{a\cdot s}{s}=a$.
. This is also the only global minimum point as a comparison with the boundary values

$\begin{array}{l}l\left(0\right)=s+\sqrt{{a}^{2}+{b}^{2}}\hfill \\ l\left(a\right)=\sqrt{{a}^{2}+{s}^{2}}+b\hfill \end{array}$

will show. From

$\begin{array}{ll}l\left(\frac{a\cdot s}{b+s}\right)\phantom{\rule{0.4em}{0ex}}\hfill & =\sqrt{\frac{{a}^{2}{s}^{2}}{{\left(b+s\right)}^{2}}+{s}^{2}}+\sqrt{{\left(a-\frac{a\cdot s}{b+s}\right)}^{2}+{b}^{2}}\hfill \\ \hfill & =\frac{s}{b+s}\sqrt{{a}^{2}+{\left(b+s\right)}^{2}}+\frac{b}{b+s}\sqrt{{a}^{2}+{\left(b+s\right)}^{2}}\hfill \\ \hfill & =\sqrt{{a}^{2}+{\left(b+s\right)}^{2}}\hfill \end{array}$

and the estimates $b=\sqrt{{b}^{2}}<\sqrt{{a}^{2}+{b}^{2}}$ and $s=\sqrt{{s}^{2}}<\sqrt{{a}^{2}+{s}^{2}}$ respectively we get

$\begin{array}{ll}1.\hfill & {a}^{2}+{\left(b+s\right)}^{2}<{a}^{2}+{b}^{2}+{s}^{2}+2s\sqrt{{a}^{2}+{b}^{2}}={\left(s+\sqrt{{a}^{2}+{b}^{2}}\right)}^{2}\hfill \\ \text{ }\text{ }⇒\text{ }\hfill & \sqrt{{a}^{2}+{\left(b+s\right)}^{2}}

All in all we thus have:

the path has a minimal length$\text{ }⇔\text{ }x=\frac{a\cdot s}{b+s}$

which finally proves the assertion [7.11.2] due to . 