In order to apply mathematical principles to the study of physical phenomena we need to agree on some foundational assumptions which seem to govern the behavior of nature. A fundamental tenet that scientists use is this: Nature is lazy! That is, nature optimizes. It tries to do things in the most efficient (optimal) manner possible. The most common form such Optimization Problems take is to find the value of the independent variable (often denoted \(x\)) which makes the dependent variable (often denoted \(y\)) as large or as small as possible.
To illustrate the usefulness of this principle consider a beam of light bouncing off of a flat mirror. Empirical evidence (that is, evidence gained from measuring the angles) suggests that the angle of incidence is equal to the angle of reflection. However, no matter how many experiments we run there is always the chance that there is some configuration of lights and mirrors we’ve missed. A configuration where the angles are not equal.
Let’s try a thought experiment.
We want to discover what path a beam of light would travel between the points \(A\) and \(B\) in the figure below assuming that it first reflects off of the line \(m\) (\(m\) for mirror, not slope).
From our assumption that Nature is Lazy, it is clear that our task is to find the shortest path from \(A\) to \(m\) to \(B\text{.}\) That is, we have an Optimization Problem.
Begin by reflecting \(B\) across \(m\) in our imagination and denote this reflection by \(B_R\text{.}\)
Clearly if the mirror were not present then the solid line from \(A\) to \(B_R\) in the figure above will be shorter than any other path from \(A\) to \(B_R\text{.}\) For example it is shorter than the dotted path from \(A\) to \(B_R\text{.}\)
Next we reflect \(B_R\) back across \(m\) so that it falls onto \(B\text{,}\) and we reflect those portions of the solid and dotted paths that are below line \(m\) in the same way.
Problem3.3.1.
Suppose we label the angles in the previous diagram as follows:
(a)
Explain why \(\angle\gamma=\angle\alpha\text{.}\)
(b)
Explain why \(\angle\gamma=\angle\beta\text{.}\)
(c)
And finally, conclude that \(\angle\alpha=\angle\beta\text{.}\)
Thus the angle of incidence (\(\alpha\)) will equal the angle of reflection (\(\beta\)) provided light adheres to the Nature is Lazy principle and really does travel by the shortest possible path.
Problem3.3.2.
If a spotlight is mounted at point \(A\text{,}\) how far from point \(C\) must point \(D\) be so that the light will be reflected to point \(B\text{?}\) Justify your answer.
Problem3.3.3.
As long as it is not spinning sideways a billiard ball will bounce just like a beam of light. In the sketch below what must \(x\) be to sink the ball into the side pocket at \(S\text{?}\) Make sure that you explain how you obtained your answer.
Another optimization problem that we can solve without Calculus (this time we’ll use Algebra) is the following.
Example3.3.4.Maximizing the Area of a Rectangle.
Out of all rectangles with a given fixed perimeter, which one encompasses the greatest area?
A reasonable guess is that the correct shape is a square. But a guess, no matter how much it feels right, is just a guess. We need a convincing demonstration.
To see that this actually is the correct shape, consider a square whose side is \(s\text{.}\) If \(s\) is the length of one side then the perimeter of our square is (the fixed value) \(4s\) and the area is \(s^2\) (also fixed). Suppose we now increase the length of one of the two pairs of opposite sides by \(x\text{.}\) To maintain the same perimeter, we must also decrease the length of the other pair of sides by the same \(x\text{.}\) The area of this new rectangle is
Thus the \(s\times s\) square has a larger area than the \((s-x) \times (s+x)\) rectangle with the same perimeter. Moreover, since we didn’t specify a value for \(x\) our conclusion holds for all possible values of \(x\text{.}\) That is, every rectangle with perimeter \(4s\) has a smaller area than the square with perimeter \(4s\text{.}\)
DIGRESSION: Variables, Constants, and Functions.
It is easy to have the impression that using a letter, like \(s\text{,}\) to represent quantity automatically means that the quantity can vary — that it is “variable.” This is not true. There are three situations in which a letter is used to represent a given quantity.
Variables:
For example, the area of a square is given by \(A=x^2,\) where \(x\) is the length of one side. If we think of \(x\) as a variable — if it can take on any positive value — then this expression gives us the area of all possible squares.
Constants:
Sometimes letters are used to designate unchanging quantities — constants (also called parameters) — when the actual value of the constant is irrelevant. For example, the area of all rectangles where one side is some fixed value, say \(c\text{,}\) but the other varies is given by \(A = c\cdot
x,\) where \(x\) is variable as before. Since we don’t care what the size of the fixed side is we don’t choose a specific length for it. We just let \(c\) represent that fixed size, whatever it is. The crucial fact here is that we have a rectangle, not what size the rectangle is.
The difference between variables and constants is mostly in the way we think about them. Constants are fixed, but unknown. Variables are not fixed.
Functions:
Sometimes a varying quantity depends on the values of one or more other varying quantities. In that case the dependent variable is said to be a function of the independent variable. We saw this above when we said the area of a rectangle with a fixed side length, \(c\text{,}\) was
\begin{equation*}
A=c\cdot x.
\end{equation*}
Most often we would write this as
\begin{equation*}
A(x)=c\cdot x.
\end{equation*}
The “\((x)\)” part of “\(A(x)\)” just tells us that that the dependent variable \(A\) depends on is the independent variable \(x\text{.}\)
If you are inclined to be persnickety, you can make a reasonable argument that if \(A(x)=cx\) then \(A\) depends on both \(c\) and \(x\text{.}\) And you’d be right. But the tradition is to only put variables inside the parentheses, not parameters.
These are simple distinctions now, but it will be useful for you to get in the habit of making sure it is clear in your mind what is variable, what is constant, and which variables are functions of other variables. As the problems get more complex and we need several variables, parameters, and functions, just untangling all of this can become problematic. Keep in mind that the distinction is purely in how we choose to “think about” a given symbol, not in how we manipulate it algebraically.
The following problem is a variation on Example 3.3.4 and can be solved similarly.
Problem3.3.5.
Out of all rectangles with a fixed perimeter, which one has the shortest diagonal? Justify your answer.
As we mentioned above, Problem 3.3.5 can be done in a manner similar to Example 3.3.4. But what about the following?
Example3.3.6.
Consider all square based boxes with a fixed surface area \(S\text{.}\) Does the cube enclose the largest volume?
To do this problem in the same manner as Example 3.3.4, observe that a cube with side \(s\) would have a volume of \(s^3\) and a surface area of \(6s^2\text{.}\) Suppose we change the lengths on the base from \(s\) to \(s+x\text{.}\) Notice that if \(x\gt0\) then we will be increasing the lengths on the base and if \(x\lt0\) then we will be decreasing the lengths. The only restriction is that \(0\lt s+x\) so as to be a length. This says that \(-s\lt x\lt\infty\text{.}\) (In theory, we can make the lengths of the base as long as we wish.) To maintain a surface area of \(6s^2\text{,}\) we will need to adjust our height appropriately. Let’s call this new height \(h\text{.}\)
Problem3.3.7.
Proceed as in Example 3.3.6 to show that in order to maintain a surface area of \(6s^2\text{,}\)\(h\) must be \(h=\frac{s^2-sx-\frac{x^2}{2}}{s+x}\text{.}\) Use this to show that the volume of this new box is
How does this address the original question about the cube having the largest volume?
Hint.
Why did we write \(V\) in this rather peculiar way?
Problem 3.3.7 shows that proceeding in this ad hoc manner is becoming more difficult. Difficulties of this kind are what prompted mathematicians in the seventeenth century to search for more systematic methods for handling such optimization. These techniques ultimately led to the invention of Calculus. We’ll come back to the this problem later in this chapter.
