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Math 220-1 Single-variable differential calculus Kursobjekt

Aaron Greicius

Section 26 Applied optimization

Example 26.1. Box of maximum volume.

A square piece of cardboard of dimension \(2\) meters is made into a box by cutting squares of equal dimension out of each corner and folding up the sides. (See Figure 26.2.) Find the dimension of the cutout squares that maximizes the volume of the resulting box.
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A square to a box
Figure 26.2. A box from a square
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Solution.
More detailed square-to-box diagram
Figure 26.3. More detailed square-to-box diagram
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Let \(x\) be the dimension of the cutout squares. The resulting box would have height \(x\text{,}\) and width and length both equal to \(2-2x\text{.}\) Its volume would thus be
\begin{equation*} V=f(x)=x(2-2x)(2-2x)=4x(x-1)^2\text{.} \end{equation*}
Since the cutout square dimension \(x\) can range from \(0\) to \(1\text{,}\) we see that we wish to find the minimal value of \(V=f(x)\) on the interval \([0,1]\text{.}\) We have reduced the question to an extreme value theorem problem!
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Following Procedure 20.13, we compute
\begin{align*} f'(x) \amp =4((x-1)^2+2x(x-1))\\ \amp = 4(x-1)(3x-1)\text{.} \end{align*}
From this factored form we see easily that the critical points of \(f\) are \(x=1\) and \(x=\frac{1}{3}\text{.}\) Now evaluate \(f\) at the endpoints of \([0,1]\) and the critical points:
\begin{align*} f(0) \amp = 0\\ f(1/3) \amp = 4\cdot \frac{1}{3}(-2/3)^2\\ \amp = \frac{16}{9}\\ f(1) \amp = 0\text{.} \end{align*}
We conclude that choosing the dimension of the cutout square to be \(x=1/3\) meters results in a box of largest possible volume.
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Example 26.5. Minimal fence perimeter.

Farmer Dudley is building a rectangular pen for his iguanas. He will use the 50 meter long side of his barn as one side of the pen, and will construct fencing for the remaining three sides of the pen. The pen must have a total area of 200 m\(^2\text{.}\) What is the minimum length of fencing Dudley must build to create an iguana pen matching these specifications.
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Solution.
Dudley’s iguana pen
Figure 26.6. Dudley’s iguana pen
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Let \(x\) and \(y\) be the dimensions of the pen, where \(x\) is the length of the barn that Dudley makes use of. Since the area \(A=xy\) of the pen must be 200 m\(^2\text{,}\) we see that \(x\) and \(y\) must satisfy
\begin{equation} A=200=xy\text{.}\tag{26.1} \end{equation}
(This is often called a constraint equation.) The quantity that Dudley wishes to minimize is the length \(L\) of fence he must build, which is
\begin{equation*} L=x+2y\text{.} \end{equation*}
Our constraint equation (26.1) implies that \(y=200/x\text{,}\) and hence that
\begin{equation*} L=f(x)=x+\frac{400}{x}\text{.} \end{equation*}
Lastly, the constraint equation implies that \(x\) cannot be equal to zero; and since Dudley is using the barn side as one edge of his pen, we must have \(x\leq 50\text{.}\) Thus, we wish to find the minimal value of \(L=f(x)\) on the domain \(D=(0,50]\text{.}\)
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We first find the critical points of \(f\) on this domain, since they are potential points where \(f\) attains a local minimum value. We solve
\begin{align*} f'(x)\amp = 0 \\ 1-\frac{400}{x^2} \amp = 0\\ x^2 \amp =400\\ x \amp =\pm 20\text{.} \end{align*}
Thus the only critical point of \(f\) lying in \(D\) is \(x=20\text{.}\) The question now, is whether \(f(20)\) is the absolute minimum value of \(f\) on \(D\text{.}\) Note that we cannot apply Procedure 20.13 to answer this question, since our domain \(D\) is not a finite closed interval! Instead we look at the sign diagram of \(f'\text{.}\)
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Sign diagram for derivative of f
From this sign diagram we conclude that \(f\) is decreasing on the interval \((0,20]\) and increasing on the interval \([20,50]\text{.}\) It follows that
\begin{equation*} f(20)=20+20=40 \end{equation*}
must be the absolute minimum value of \(f\) on \(D\text{,}\) since for any other \(x\in D\text{,}\) we have \(f(x)> f(20)\text{.}\) We conclude that minimal length of fencing Dudley can build to make his pen is \(40\) meters.
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We conclude with a graph of \(f\text{,}\) the simple shape of which makes more tangible our logic above in arguing that \(f\) attains its absolute minimum value at \(x=20\text{.}\)
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Graph of f
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As we saw in the last example it is often the case that the domain \(D\) of the function we wish to optimize is not of the simple form \([a,b]\text{,}\) where Procedure 20.13 applies. In Example 26.5, we were optimizing a function on the interval \((0,50]\text{,}\) and had to rely on our wits a bit to convince ourselves that we have found an absolute extreme value. The following generalization of the extreme value theorem gives a more systematic procedure for finding absolute extreme values in such situations.
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Example 26.8. Minimal fence perimeter (reprise).

Let’s use Procedure 26.7 to complete the last steps of our argument in Example 26.5. We were trying to find the absolute minimum value of \(f(x)=x+\frac{400}{x}\) on the half-open interval \(I=(0,50]\text{,}\) and had determined that \(x=20\) was the sole critical point of \(f\text{.}\) Procedure 26.7 instructs us to evaluate \(f\) at \(x=20\) and \(x=50\text{,}\) and compute the limit at \(0\text{:}\)
\begin{align*} \lim\limits_{x\to 0^+ }f(x) \amp = \lim\limits_{x\to 0^+ }x+\frac{400}{x}=\infty \amp (\text{type } 0+\infty) \\ f(20) \amp = 20+20=40\\ f(50)\amp =50+400/50=58 \text{.} \end{align*}
We conclude that \(f(20)=40\) is the absolute minimum value of \(f\) on \((0,50]\) (and that \(f\) has no absolute maximum value).
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Example 26.9. Closest point on parabola.

Find the \(Q\) on the parabola \(y=1-x^2\) whose distance to \(P=(-3,1)\) is the smallest possible. The distance between two points \(P_1=(x_1,y_1)\) and \(P_2=(x_2,y_2)\) in \(\R^2\) is defined as
\begin{equation*} d(P_1,P_2)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\text{.} \end{equation*}
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Solution.
Diagram of point P and parabola
Figure 26.10. Distance between \(P\) and points on parabola \(y=1-x^2\)
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The distance from \(P\) to an arbitrary point
\begin{equation*} Q=(x,y)=(x,1-x^2) \end{equation*}
on \(\mathcal{C}\) is given by
\begin{equation*} d=f(x)=\sqrt{(-3-x)^2+(1-(1-x^2))^2}=\sqrt{(x+3)^2+x^4}\text{.} \end{equation*}
Since the \(x\)-coordinate of \(Q\) can be any real number, we wish to find the minimum value of \(f\) on \(\R=(-\infty, \infty)\text{.}\) We follow Procedure 26.7.
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Since \(f\) is differentiable everywhere, its critical points are solutions to \(f'(x)=0\text{.}\) We solve:
\begin{align*} f'(x) \amp = 0\\ \frac{2(x+3)+4x^3}{2\sqrt{(x+3)^2+x^4}} \amp = 0 \\ 4x^3+2x+6 \amp =0\text{,} \end{align*}
where the last step follows from the fact that a quotient is equal to zero precisely when its numerator is equal to zero.
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Let \(g(x)=4x^3+2x+6\text{.}\) In general it is not so easy to find roots of a cubic polynomial. However, recall the fact that if \(g\) has an integer root, it must be a divisor of \(6\text{.}\) Trying a few divisors (\(\pm 1, \pm 2, \pm 3, \pm 6\)), we easily see that \(g(-1)=0\text{,}\) and hence that \((x+1)\) is a factor of \(g(x)\text{,}\) Using polynomial long division, we see that
\begin{equation*} g(x)=(x+1)(4x^2-4x-3). \end{equation*}
Furthermore, using the quadratic formula, we see that the \(4x^2-4x-3\) has no real roots. Thus the root of \(g\) (and only critical point of \(f\)) is \(x=-1\text{.}\)
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According to Procedure 26.7, to investigate extreme values of \(f\) on \((-\infty, \infty)\) we should evaluate \(f\) at \(x=-1\) and compute the limits of \(f\) at \(\pm \infty\text{:}\)
\begin{align*} f(-1) \amp = \sqrt{4+1}=\sqrt{5}\\ \lim\limits_{x\to \pm \infty}f(x) \amp =\lim\limits_{x\to \pm \infty}\sqrt{(x+3)^2+x^4}=\infty \amp (\text{type } \sqrt{\infty})\text{,} \end{align*}
where the limit at infinity computations follow from the fact that \((x+3)^2+x^4\to \infty\) as \(x\to \pm \infty\text{.}\)
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We conclude that \(f(-1)=\sqrt{5}\) is the absolute minimum value of \(f\text{,}\) and thus that \(Q=(-1,0)\) is the point on \(\mathcal{C}\) closest to \(P\text{.}\)
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