Section9.2Preliminaries: Some Simple Optimizations
Our first goal is to build a set of tools for optimization. By the end of this chapter we want to be able to rely more on the tools of Calculus and less on intuition to solve optimization problems. This does not mean that you should stop using your intuition. But by the end of this chapter you will have a set of tools that will allow you to either confirm your instincts, or refute them.
To begin we will consider problems that allow us to use our intuition. As a result many of the examples in this section can be solved by more elementary methods than Calculus. If you can see a more fundamental way to solve these problems and examples, that’s great! Do that. But keep your learning goal in mind. We’re interested in these examples as a means to elucidate the subtleties inherent in optimization problems. These problems are a means to that end, not an end in themselves. Don’t just “get the right answer” and stop. Learn the process and look for where difficulties could arise in a more complex situations.
Example9.2.1.Constructing A Square on a Line.
What is the area of the smallest possible square having one corner at the point \((2,3)\) and an adjacent corner on the line
\begin{equation}
5x+3y=2.\tag{9.1}
\end{equation}
The diagram below is a visualization of this problem.
We can imagine the blue square growing and shrinking as we move the point \((a,b)\) along the line. It seems clear that the square having the least area is the one pictured in black (when \((a,b)\) is at \((x,y)\text{.}\) It is worth taking a moment to notice this explicitly because not all optimization problems are solvable. If we had asked for the largest possible square there would be no solution because the blue square in our diagram can always be make larger by moving the point \((a,b)\) further to the left on the graph of equation (9.1).
But a convincing picture is not a solution. To complete the solution we’ll need to compute the coordinates of the point \((x,y)\) in our sketch and compute the minimal area.
In any optimization problem the first question to ask is, “What do we need to optimize?” In other words, “What is our objective function?” For this problem the length of the line segment between the point \((2,3)\) and any point \((x,y)\) is given by the Distance Formula: \(\sqrt{(x-2)^2+(y-3)^2}\text{.}\) Therefore the area of the square constructed from this line segment is given by the function:
So our objective function is \(A(x)\text{.}\) Since the point \((x,y)\) is constrained to lie on the line \(5x+3y=2\text{,}\)equation (9.1), is our constraint.
Notice that we’ve abused our notation a bit. Although the notation \(A(x)\) indicates that \(A\) depends on \(x\) alone, we’ve given \(A\) in terms of both \(x\) and \(y\text{.}\) But solving equation (9.1) gives \(y=\frac13(2-5x)\) so it should be clear that \(A\) really depends only on \(x\text{.}\)
We’ve seen visually that this problem has a solution and Fermat’s Theorem 9.1.1 tells us that the derivative of our objective function will be zero at that solution. So we differentiate both sides of
\begin{equation*}
A =(x-2)^2+(y-3)^2
\end{equation*}
But recall that \(x\) and \(y\) are constrained by the equation (9.1) from which we see that \(\textcolor{red}{\dx{y}= -\frac{5}{3}\dx{x}}\text{.}\) Therefore
So we have found that the minimal square occurs when \(x=-\frac12\text{,}\) and \(y=\frac32\) and it feels like we’re done, doesn’t it?
We’re not.
The question asked was “What is the area of the smallest possible square?” To find the area we need to put \((x,y)=(-1/2, 3/2)\) into our area formula which yields
Always remember to check that you have actually answered the question which was asked before you stop.
Problem9.2.2.
(a)
Do you see that the point \((x,y)\) which minimizes the area of our square in Example 9.2.1 also minimizes the length of the line segment from \((2,3)\) to \((x,y)\text{?}\) What objective function would we get if we tried to minimize the length of the segment instead of the area of the square. Does this change the problem? If so, how?
(b)
This example specified that \((2,3)\) and \((x,y)\) were adjacent corners of the square. What would the objective function look like if those points were diagonally opposite each other. Does this change the problem? If so, how?
Problem9.2.3.
It seems fairly intuitively clear that the minimal square in Example 9.2.1 occurs when the line through \((2,3)\) is orthogonal to the graph of the line \(5x+3y=2\text{.}\)
(a)
Verify that our intuition is correct.
(b)
Does this remain true if the point on the line and \((2,3)\) are not adjacent corners of the square?
We have a few general comments regarding Example 9.2.1.
The first step in any optimization problem is to identify the objective function. Sometimes more than one will work. In that case use the simplest one, but without an objective function we can’t even get started.
We need to identify constraint(s) as well, if any exist. (Not all optimization problems will have constraints. Some will have more than one.) We may not need these immediately, but we will need them sooner or later.
It was intuitively clear from our sketch that there actually is a minimal square in this example; that a solution exists. This will be true often enough that it will be easy to start believing that you can rely on your intuition alone to solve optimization problems. This is a mistake. Intuition is important and we encourage you to use it, especially when you first start thinking about a new problem. But intuition is based on past experience. It is a way of saying that you expect today to go pretty much the same way yesterday did. Most of the time this is true, but you never know when things will go awry so you must always verify that your intuition is correct by some other means. Calculus, in this case.
Most of the heavy lifting in this problem was done in the Algebra, not the Calculus. This will be true of many of the mathematics problems you will encounter from now on. There is a strong tendency to think of algebraic manipulations as “just algebra” and to be careless with them. This is a mistake. Every step toward solving a problem is just as important as every other step, whether it is Algebra, Calculus, Geometry, Arithmetic, or just “thinking about the problem.” Every step must be successfully completed. Strive to be skillful, and careful, with all of your tools.
We were very careful to write out, in detail, every algebraic and arithmetic step in this problem but if we were to do this for every problem this text would become too long to be useful. We don’t want that to happen so we will regularly do several elementary algebraic steps at once. But “elementary” does not mean “easy.” You may not be able to follow all of the steps in your head, especially at first. When you do not see how to get from one line of computation to the next you may be tempted to just skip it.
Don’t. Do. That.
Keep a writing pad handy at all times. If there is any question of understanding a given computation take out that pad and do the computation yourself. If you are unable to do the computation yourself, ask for help. Ask a classmate, a friend, your instructor, whomever. Ask for help when you need it. If you do this habitually you will find that following the steps gets easier over time. If you do not it will remain difficult forever. We’re serious. Do all elementary computations yourself! It is your responsibility to make sure you understand the computations we show you.
Example9.2.4.Constructing A Square on a Circle.
We want to find the area of the largest and the smallest squares that can be constructed with one corner at the point \((2,3)\) and an adjacent corner on the unit circle: \(x^2+y^2=1\text{.}\)
The solution of this problem is very clear once it has been sketched. We will solve it again analytically below.
Drill9.2.5.
The purpose of this drill is for you to see the solution of Example 9.2.4 intuitively.
Sketch the unit circle, plot the point \((2,3)\text{,}\) and draw the smallest and largest squares you can which have one corner at the point \((2,3)\) and an adjacent corner on the unit circle: \(x^2+y^2=1\text{.}\)
Explain how you can tell from your sketch that that there is both a maximal (largest area) and a minimal (smallest area) square.
Explain how you can tell from your sketch, that the second corner of the maximal and minimal squares are at the intersection points of the line \(y=\frac32x\) and the unit circle.
Show that the areas of the maximal square is \(A\approx 21.2\) units and that the area of the minimal square is \(A\approx6.8\) units.
The solution in Drill 9.2.5 relies heavily on the intuition gained from sketching the problem. As we have said many times you should use your intuition but not rely on it. To confirm our solution in Drill 9.2.5 using Calculus tools we will start at the beginning and solve this problem again, relying as much as possible on our computations. Eventually, we would like to rely exclusively on our computations, but this is not quite possible yet.
We have the same objective function as in Example 9.2.1:
In order to use Fermat’s Theorem 9.1.1 we must first know that one maximum and one minimum exists. No computation we’ve done guarantees the existence of an extremum, let alone its location. However, as a result of the work you did to solve Drill 9.2.5 we do know, intuitively, that there are two optimal solutions: one maximal and one minimal. Therefore Fermat’s Theorem does guarantee that at each of these (wherever they are) \(\dfdx{A}{x}=0\text{.}\)
We proceed as just discussed. Since we know (from Drill 9.2.5) that both a maximal and a minimal solution exist we set \(\dfdx{A}{x}= 2\left(-2+\frac{3x}{y}\right)=0\text{.}\) Solving this yields
\begin{equation}
3x=2y,\tag{9.7}
\end{equation}
and since \(y=\pm\sqrt{1-x^2}\) (from our constraint, equation (9.5)) we have
The square root symbol, and the plus–or–minus (\(\pm\)) make equation (9.8) seem a little scary at first but it is really just a quadratic equation in disguise. Squaring both sides of equation (9.8) gives \(9x^2=4(1-x^2)\) or
\begin{equation}
13x^2=4.\tag{9.9}
\end{equation}
So we have two solutions: \(x=\frac{2}{\sqrt{13}}\) and \(x=-\frac{2}{\sqrt{13}}\text{.}\) From Drill 9.2.5 we know that these must correspond to the maximal and minimal solutions. Fermat’s Theorem 9.1.1 doesn’t tell us which is which but all we need to do is distinguish between them is evaluate \(A(x)\) for each.
Drill9.2.6.
Show that the area of the maximal square is approximately \(21.2\) units. Which value of \(x\) does this correspond to?
Show that the area of the minimal square is approximately \(6.8\) units. Which value of \(x\) does this correspond to?
Problem9.2.7.
There are two extremal squares with one corner at \((2,3)\) and an adjacent corner on the ellipse:
In solving this problem it is possible that you will encounter a sub–problem that cannot be easily solved algebraically. In that case use Newton’s Method or some computational software as needed to find approximations.
Since a circle is a special case of an ellipse this problem is a slight generalization of Drill 9.2.5 so the algebra is a bit trickier.
(a)
Draw a sketch of this problem. Explain how you can tell from the sketch that there is one maximal, and one minimal square.
(b)
The objective function for this problem is, once again:
Show that \(\dfdx{A}{x}=-6x-4+\frac{24x}{y}\text{.}\)
(e)
Show that if we choose \(y=2\sqrt{2-x^2}\) then the only root of \(\dfdx{A}{x}=0\) is \(x\approx0.86\text{.}\)
(f)
Show that if we choose \(y=-2\sqrt{2-x^2}\) then the only root of \(\dfdx{A}{x}=0\) is \(x\approx-0.28\text{.}\)
(g)
Find (approximately) the area of the square associated with both of these points. Which one is the maximal and which is the minimal square?
Problem9.2.8.Constructing a Square on an Ellipse.
Here is a sketch of the optimal squares in Problem 9.2.7. Notice that for both the maximal and the minimal square one of the sides appears to be orthogonal to (the line tangent to) the ellipse.
(a)
Show that they actually are orthogonal.
(b)
It is generally true that the shortest (or longest) line segment from a point \((a,b)\) to a curve, \(C\text{,}\) will be orthogonal to the line tangent to the curve at their point of intersection. The figure below shows an arbitrary, differentiable curve, \(C\text{,}\) and an arbitrary point \((a,b)\text{.}\)
Suppose that the area of a square with one corner at \((a,b)\) and an adjacent corner on the curve at \((x,y)\) is optimal (minimal or maximal) at \((x_0, y_0)\text{.}\) Show that the line joining \((a,b)\) and \((x_0,y_0)\) is orthogonal to the tangent line of \(C\) at \((x_0,y_0)\) . Try to solve this problem completely on your own, but if you get stuck here is an outline you can follow:
Find the objective function, \(A(x)\text{.}\) (This will be the area of the square on the line from \((a,b)\) to \((x,y)\text{.}\))
Use the fact that \(\mineval{\dfdx{A}{x}}{(x_0,y_0)}=0\) to find the slope of \(C\) at \((x_0,y_0)\text{.}\)
Find the slope of the line from \((a,b)\) to \((x_0,y_0)\text{.}\)
Compare the two slopes.
What happens if the tangent line is horizontal? If it is vertical?
Example9.2.9.Maximizing the Area of a Rectangle, Redux: Differentials vs. Derivatives.
Take a moment to reconsider Example 3.3.4. Is it clear to you that there must be such a maximal rectangle? If so then, as before, setting the derivative of the objective function equal to zero and solving for the variable will provide all of the values of the variable where the maximal rectangle might occur. If we’re lucky there will only be a few such and we can check them one at a time.
Finding the objective function in this case is easy. The area \(A\text{,}\) of any rectangle is its length, \(l\text{,}\) times its width, \(w\text{,}\) so
and is constrained to be a constant (fixed) value so equation (9.11) is the constraint. Throughout this example keep in mind that \(P\) is a fixed constant, not a function of \(l\) and \(w\text{.}\)
At this point we can complete the problem using Calculus in either of two equivalent ways.
Using Differentials:
In our opinion this is how Leibniz intended for his Calculus to be used so we tend to favor it.
Compute \(\dx{A} = l\dx{w}+w\dx{l}\text{,}\) from the objective function.
Differentiate the constraint, giving the differential form of the constraint: \(2\dx{l}+2\dx{w}=0.\)
Set \(\dx{A}=0\) and solve this equation using the constraint and its differential form as needed. In this problem we have \(\dx{l}=-\dx{w}\) so
Since \(\dx{w}\neq0\) we have \(l=w\) and thus the square encloses the most area.
Using Derivatives:
The modern approach emphasizes the use of functions and derivatives rather than curves and differentials.
Solve the constraint for either \(w\) or \(l\) in terms of the other. For example, \(l=\frac12(P-2w).\)
Substitute this into the objective function to rewrite it in terms of a single variable:
\begin{equation*}
A(w)=\frac12(P-2w)\cdot w = \frac12(Pw-w^2).
\end{equation*}
Now think of the objective function, \(A(w)\text{,}\) as an abstract function and find its maximum point by setting \(\dfdx{A}{w}=0\) and solving for \(w\text{.}\)
so \(w=\frac{P}{4}\text{.}\) Putting this back into our constraint gives \(P=2\cdot\frac{P}{4}+2l\) we see that \(l\) is also equal to \(\frac{P}{4}\text{.}\) Since \(l=w=\frac{P}{4}\) the square encloses the most area.
As we said, the two approaches are completely equivalent. Use whichever feels most comfortable to you.
In Chapter 3 we used Fermat’s Method of Adequality to solve the following problem which is closely related to Example 9.2.9,
Problem9.2.10.
Find the dimensions of the rectangle having the smallest perimeter, given that the area is fixed. Since we are still relying on Fermat’s Theorem, the first step is to convince yourself that this problem has a solution. Once you have done that, use both of the approaches outlined in Example 9.2.9 to complete the solution. Which method do you prefer?
Problem9.2.11.
Consider all lines connecting the point \((-4,-3)\) to an arbitrary point \((x,y)\text{,}\) on the unit circle as pictured. There is clearly one point where the slope of the associated line is maximum and another point where the slope of the associated line is minimum. These are the outermost lines depicted in the figure. Find an equation for each of these lines. If you aren’t sure how to begin use the following steps.
(a)
Find an appropriate objective function for this problem.
(b)
Find the constraint for this problem.
(c)
Differentiate both your objective function and your constraint to show that the points where the maximum and minimum slope occur must satisfy
Solve equation (9.12) and the constraint simultaneously to find the equations of the lines with maximal and minimal slope. Verify your answer by graphing both lines and the unit circle.
Drill9.2.12.
Sketch the graphs of \(y=e^x\) and \(y=\ln(x)\) and convince yourself that there is a point on each graph which is closest to the origin.
Find the approximate coordinates of the point on the curve \(y=e^x\) that is closest to the origin.
Find the approximate coordinates of the point on the curve \(y=\ln(x)\) that is closest to the origin.
Are the solutions of parts (a) and (b) related? Explain.
Drill9.2.13.
Suppose a point is moving in a plane so that its \((x,y)\) coordinates are given by:
\begin{equation*}
P(t)=
\ParamEqThree {t-1} {e^t} {-\infty\lt t \lt \infty}
\end{equation*}
Find the value of \(t\) where the point, \(P(t)\text{,}\) is closest to the origin. Is it clear that there is such a point?
Find the \((x,y)\) coordinates of the point’s location when it is closest to the origin.
Problem9.2.14.Find the Pattern.
(a)
Of all the rectangles that can be inscribed in the unit circle show that the one with the maximum area is a square.
(b)
Of all the rectangles that can be inscribed in the ellipse \(x^2+\frac{y^2}{9}=1\) show that the one with the maximum area is three times taller than it is wide.
(c)
Of all the rectangles that can be inscribed in the ellipse \(\frac{x^2}{16}+y^2=1\) show that the one with the maximum area is four times wider than it is tall.
(d)
Of all the rectangles that can be inscribed in the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) show that the one with the maximum area has a height to width ratio of \(\frac{b}{a}\text{.}\)
DIGRESSION: Triangles With Fixed Perimeters.
In Examples 3.3.4 and Example 9.2.9 we saw that of all rectangles with fixed perimeter, the square (an equilateral rectangle) has the largest area. Is this also true of triangles? More precisely, does an equilateral triangle have the largest area out of all triangles with a fixed perimeter? This may sound like a simpler problem but it is actually a bit harder.
Here’s why. With a rectangle each set of parallel sides varied together so we only had two independent parameters to deal with. For triangles all three sides vary independently, giving us three independently varying parameters, which is inherently more complex than two.
How might we address this question?
A standard ploy for solving difficult problems is to first try a simpler “adjacent” problem. In this case to create an “adjacent” problem we hold one side fixed and only allow two sides of our triangle to vary. We’ll show that when the area of the triangle is maximal the two varying sides must be the same length. We want to use this adjacent problem to gain some insight into how to solve our actual problem.
So our adjacent problem is this: Given all triangles with a fixed base and a fixed perimeter, does an isosceles triangle have the largest area? The sketch below has all of the pertinent parameters labeled.
Figure9.2.15.
If we hold \(b\) fixed then minimizing the area, \(A=\frac12bh\text{,}\) is equivalent to maximizing \(h\text{.}\) So the length of \(h\) is the natural objective function for this problem. Also if \(h\) is at its maximum then by Fermat’s Theorem 9.1.1, \(\dx{h}=0\text{.}\) Notice that there are two right triangles in our sketch and \(h\) is the height of both of them.
Problem9.2.16.
(a)
Use the sketch in Figure 9.2.15 to express \(h\) as a function of \(x\) and \(z\) and show that when \(h\) is maximal \(x\dx{x} - z\dx{z}=0\text{.}\)
(b)
Express \(h\) as a function of \(y\) and \(w\) and show that when \(h\) is maximal \(y\dx{y}-w\dx{w}=0\text{.}\)
(c)
Use the fact that \(b\) is constant to find an equation relating \(\dx{z}\) and \(\dx{w}\text{.}\)
(d)
Use the fact that the perimeter \(P\) is constant to find an equation relating \(\dx{x}\) and \(\dx{y}\text{.}\)
(e)
Use the results from parts (a), (b), (c), and (d) to show that \(h\) is maximized when
Use part (e) to conclude that the base angles, \(\alpha\) and \(\beta\text{,}\) must be congruent and thus \(x=y\text{.}\)
To show that the maximal area of an arbitrary triangle with a fixed perimeter must be equilateral we do not need Calculus, only reasoning. An example of the kind of reasoning we need appears in the Sherlock Holmes short story The Adventure of the Beryl Coronet. Early in the story Holmes makes the following statement to to Watson.
Principle9.2.17.Sherlock Holmes’ Maxim.
“It is an old maxim of mine that when you have excluded the impossible, whatever remains, however improbable, must be the truth.”
Our solution will follow Holmes’ lead: We will eliminate what is impossible leaving only the truth behind.
Problem9.2.18.The Largest Triangle with a Fixed Perimeter is Equilateral.
Observe that every triangle is either equilateral, or it is not equilateral.
(a)
Suppose we have a triangle with sides \(a\text{,}\)\(b\text{,}\) and \(c\) which is not equilateral. Then at least two sides, say \(a\) and \(c\text{,}\) are not equal. Explain how Problem 9.2.16 implies that there must be another triangle with the same perimeter and a larger area.
(b)
Use Holmes’ maxim to explain why a triangle with a fixed perimeter enclosing maximal area must be equilateral.
