## 170- Quiz 9

Quiz 9 is here.

Solutions follow.

Problem 1 tells us that a rectangular piece of cardboard of dimensions $8\times 17$ is used to make an open-top box by cutting out a small square of side $x$ from each corner and bending up the sides. If $x = 2$, then the volume of the box is $2\times 4\times 13=104$. Starting from this initial guess of $x=2$, the problem asks us to use Newton’s method to find a value of $x$ for which the box has volume 100, accurate to 3 significant figures.

To solve the problem, first we observe that for any $x$, the volume of the box is $V(x)=(8-2x)(17-2x)x$, as a quick diagram would indicate. We want to solve the equation

$V(x)=100$

using Newton’s method. Since the method is designed to solve equations of the form $f(x)=0$, we take $f(x)=V(x)-100$. In the method, we begin with an initial guess $x_0$ and then compute further guesses $x_1,x_2,\dots$ by means of the formula

$\displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$

In the case that concerns us, $f(x)=(8-2x)(17-2x)x-100$ and (using the product rule) $f'(x)=-2(17-2x)x-2(8-2x)x+(8-2x)(17-2x)$.

The file below was produced using the software package Sage, which can be downloaded for free at the link. (As a side note, if you expect through your careers you will be doing a significant amount of non-trivial computations, it may be a reasonable investment of time to learn how to use one or two software packages. I like Sage personally, but of course there are many other alternatives, although not all are free.)

In the file, the value of each new guess is shown both exactly (as a fraction) and then numerically. As can be seen from the computations above, $x=2.1919157$ is a very reasonable approximation to the solution of $V(x)=100$.

Problem 2 tells us that the U.S. post office will accept a box for shipment only if the sum of the length and girth (distance around) is at most 108 in. We are asked to find the dimensions of the largest acceptable box with square front and back.

To solve this problem, we imagine a box with square front of side length $x$, and width $y$. We are told that $x+(2x+2y)\le 108$. We are also asked to maximize the volume of the box. But the volume can be easily computed by the formula $V=x^2y$. Clearly, if we fix the value of $x$ and increase the value of $y$, the volume increases. This means that we may as well suppose that $3x+2y=108$. Then $y=(108-3x)/2$ and

$V(x)=x^2(108-3x)/2.$

We need to optimize this volume subject to the restrictions that $x\ge0$ and $x\le 108/3=36$. For this, we solve $V'(x)=0$. We have

$V'(x)=2x(108-3x)/2-3x^2/2=108x-3x^2-3x^2/2$ $=108x-9x^2/2=9x(24-x)/2.$

We have that $V'(x)=0$ iff $x=0$ or $x=24$. Hence, we need to check the value of $V(x)$ when $x=0$, $x=24$, and $x=36$.

In the first and last cases, $V(x)=0$. In the second one, $V(24)=24^2\times36$. It follows that this is the maximum volume, and it is achieved when $x=24$ inches, and $y=18$ inches.