13.2 Optimization Problems.

13.8   Example. A stick of length $l$ is to be broken into four pieces of length $s, s, t$ and $t$ and the pieces are to be assembled to make a rectangle. How should $s$ and $t$ be chosen if the area of the rectangle is to be as large as possible? What is the area of this largest rectangle? Before doing the problem you should guess the answer. Your guess will probably be correct.

Let $s$ be the length of one side of the rectangle. Then $2s+2t=l$ so $${t={l\over 2}-s}$$; i.e., $t$ is a function of $s$. Let $A(s)$ be the area of a rectangle with side $s$. Then

$$A(s)=st=s\Big( {l\over 2}-s\Big)={l\over 2}s-s^2\quad{\mbox{ for } } 0\leq s\leq {l\over 2}.$$

I include the endpoints for convenience; i.e., I consider rectangles with zero area to be admissible candidates for my answer. These clearly correspond to minimum area. Now

$$A^\prime (s)={l\over 2}-2s$$

so $A$ has only one critical point, namely $${ {l\over 4}}$$, and

$$\displaystyle { A\Big( {l\over 4}\Big) = {l\over 2}{l\over 4}... ...( {l\over 4}\Big)^2={{l^2}\over {16}}=\Big( {l\over 4}\Big)^2}.$$

Since $A$ is continuous on $${\Big[ 0,{l\over 2}\Big]}$$ we know that $A$ has a maximum and a minimum, and since $A$ is differentiable on $${ \Big( 0,{l\over 2}\Big)}$$ the extreme points are a subset of $${ \Big\{ 0,{l\over 2},{l\over 4}\Big\}}$$. Since $A(0)=\displaystyle { A\Big( {l\over 2}\Big)}=0$ the maximal area is $${ \Big( {l\over 4}\Big)^2}$$; i.e., the maximal rectangle is a square. (As you probably guessed.)

This problem is solved by Euclid in completely geometrical terms [17, vol 1 page 382].

Euclid's proof when transformed from geometry to algebra becomes the following. Suppose in our problem $s\neq t$, say $s<t$. Since $${ s+t={l\over 2}}$$, it follows that $${s\leq{l\over 4}\leq t}$$ (if $s$ and $t$ were both less than $${ {l\over 4}}$$, we'd get a contradiction, and if they were both greater than $${ {l\over 4}}$$, we'd get a contradiction). Let $r$ be defined by

$$s={l\over 4}-r \mbox{ so } r\geq 0.$$

Then $${ t={l\over 2}-s={l\over 2}-\Big( {l\over 4}-r\Big) ={l\over 4}+r}$$ so

$$A(s)=st=\Big( {l\over 4}-r\Big)\Big( {l\over 4}+r\Big)=\Big({l\over 4}\Big)^2-r^2=A\Big( {l\over 4}\Big)-r^2.$$

Hence, if $r>0$, $${A(s)<A\Big( {l\over 4}\Big)}$$ and to get a maximum we must have $r=0$ and $${ s={l\over 4}}$$. This proof requires knowing the answer ahead of time (but you probably were able to guess it). In any case, Euclid's argument is special, whereas our calculus proof applies in many situations.

Quadratic polynomials can be minimized (or maximized) without calculus by completing the square. For example, we have

$$\begin{eqnarray*} A(s) &=& -s^2+{l\over 2}s \\ &=& -\Bigg( s^2-{l\over 2}s+\Big... ...)^2 \\ &=& \Big( {l\over 4}\Big)^2 - \Big( s-{l\over 4}\Big)^2. \end{eqnarray*}$$

From this we can easily see that $${A(s)\leq\Big( {l\over 4}\Big)^2}$$ for all $s$ and equality holds only if $${ s={l\over 4}}$$. This technique applies only to quadratic polynomials.

13.9   Example. Suppose I have 100 ft. of fence, and I want to fence off 3 sides of a rectangular garden, the fourth side of which lies against a wall and requires no fence (see the figure). What should the sides of the garden be if the area is to be as large as possible?

This is a straightforward problem, and in the next exercise you will do it by using calculus. Here I want to indicate how to do the problem without calculation. Imagine that the wall is a mirror, and that my fence is reflected in the wall.

When I maximize the area of a garden with a rectangle of sides $x$ and $y$, then I have maximized the area of a rectangle bounded by 200 feet of fence (on four sides) with sides $y$ and $2x$. From the previous problem the answer to this problem is a square with $y=2x=50$. Hence, the answer to my original question is $y=50$, $x=25$. Often optimization problems have solutions that can be guessed on the basis of symmetry. You should try to guess answers to these problems before doing the calculations.

13.10   Exercise. Verify my solution in the previous example by using calculus and by completing the square.

13.11   Example. I want to design a cylindrical can of radius $r$ and height $h$ with a volume of $V_0$ cubic feet ($V_0$ is a constant). How should I choose $r$ and $h$ if the amount of tin in the can is to be minimum?

Here I don't see any obvious guess to make for the answer.

I have

$$V_0= \mbox{ volume of can } =\pi r^2 h,$$

so $${h={{V_0}\over {\pi r^2}}}$$. Let $A(r)$ be the surface area of the can of radius $r$. Then

$$\begin{eqnarray*} A(r) &=& \mbox{ area of sides } +2 \mbox{ (area of top) } \\ ... ...\over {\pi r^2}} + 2 \pi r^2 \\ &=& {{2V_0}\over r} + 2\pi r^2. \end{eqnarray*}$$

The domain of $A$ is $\mbox{{\bf R}}^+$. It is clear that $${\lim_{r\to 0^+} A(r)=+\infty}$$, and $${\lim_{r\to +\infty} A(r)=+\infty}$$. Now $${A^\prime (r)=-{{2V_0}\over {r^2}}+4\pi r={{4\pi}\over {r^2}}\Big( r^3-{{V_0}\over {2\pi}}\Big)}$$. The only critical point for $A$ is $${ r=\root 3 \of { {{V_0}\over {2\pi}}}}$$ (call this number $r_0$). Then $${ A^\prime (r)={{4\pi}\over {r^2}}(r^3-r_0^3)}$$. We see that $A^\prime (r)<0$ for $r\in (0,r_0)$ and $A^\prime (r)>0$ for $r\in (r_0,\infty )$ so $A$ is decreasing on $(0,r_0]$ and $A$ is increasing on $[r_0,\infty )$ and thus $A$ has a minimum at $r_0$. The value of $h$ corresponding to $r_0$ is

$$h={{V_0}\over {\pi r_0^2}}={{V_0}\over {\pi\Big( {{V_0}\over ... ...0^{1/3}=2\Big( {{V_0^{1/3}}\over {2^{1/3}\pi^{1/3}}}\Big)=2r_0.$$

Thus the height of my can is equal to its diameter; i.e., the can will exactly fit into a cubical box.

In the following four exercises see if you can make a reasonable guess to the solutions before you use calculus to find them.

13.12   Exercise. A box (without a lid) is to be made by cutting 4 squares of side $s$ from the corners of a $12'' \times 12''$ square, and folding up the corners as indicated in the figure.

How should $s$ be chosen to make the volume of the box as large as possible?

13.13   Exercise. A rectangular box with a square bottom and no lid is to be built having a volume of 256 cubic inches. What should the dimensions be, if the total surface area of the box is to be as small as possible?

13.14   Exercise. A Find the point(s) on the parabola whose equation is $y=x^2$ that are nearest to the point $${\Big( 0,{9\over 2}\Big)}$$.

13.15   Exercise. A Let $\mbox{{\bf p}}=(0,3)$ and let $\mbox{{\bf q}} =(12,6)$. Find the point(s) r on the $x$-axis so that path from $\mbox{{\bf p}}$ to r to q is as short as possible; i.e., such that length([p r]) $+$ length([r q]) is as short as possible.

You don't need to prove that the critical point(s) you find are actually minimum points.