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What is the Largest-Volume, Open-Top, Rectangular Box You Can Make from a Sheet of Cardboard? – Exploring Polynomial Functions

A problem involving geometry, algebra and calculus made easy by

spreadsheets

Prepared for SSAC byNasser Dastrange, Buena Vista University, Storm Lake, Iowa, 50588

© The Washington Center for Improving the Quality of Undergraduate Education. All rights reserved. 2007

SSAC2007.QA154.ND1.1

Core Quantitative ConceptPolynomial functions

Supporting Quantitative ConceptsCalculus – Finding maxima and minima of polynomial functionsAlgebra – Quadratic formulaGeometry – Volume of a rectangular prismGraphing – XY scatter plot

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Introduction

A polynomial function p(x) has the standard form

p(x) = an x n + a n−1 x n−1 + … + a1 x + a0

The function is a sum of terms each consisting of a constant coefficient (ai) times the independent variable (x) raised to a non-negative integer exponent; the terms are arranged in order of decreasing exponents. The degree of the polynomial function is the value of the largest exponent (n). For example,

f(x) = 5x3 − 24x2 + 13x + 27is a third-degree polynomial function (also known as a cubic function; a second-degree polynomial function is also called a quadratic function) The constant term (27, in this case) is a coefficient times x raised to the zero power.

This module illustrates methods of finding local maxima and minima of polynomial functions. One method uses calculus, specifically the concept that the first derivative of a function is zero at local maxima and minima. The derivative of a polynomial function of degree n is another polynomial function of degree n−1. As a result, this method quickly becomes an algebra problem of finding the roots of a polynomial equation. Here is an opportunity for you to use the quadratic formula. A spreadsheet makes it easy.

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The Problem and Preview

You have a sheet of light cardboard and you want to fold it into an open-top box by cutting a square of length x from each corner and folding up the sides. The sheet of cardboard is 8 inches by 10 inches. What value of x maximizes the volume of the box?

10 inches

8 inches

xx

There are numerous ways we can answer this question.

Slide 4 directs you to a site on the Internet where you can solve the problem experimentally with an applet, and Slide 5 describes how you might determine the answer experimentally using cardboard, scissors and some granular cereal.

Slide 6-9 guide you to solve the problem using a spreadsheet to calculate the volume for numerous values of x, and Slide 10 asks you to graph the polynomial function given in the list of x and V.

Slide 10-13 guide you to solve the problem using calculus and the quadratic formula.

Slide 14 asks you to build a spreadsheet to find the roots of a quadratic equation assuming you don’t remember the quadratic formula.

Slide 15 has the end-of-module assignments.

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Using the Internet

You can try to find something on the Internet. For example, you might Google on “three dimensional box” and find a nifty animation titled “Three dimensional box applet: Working with volume.”

10 inches

8 inches

xx

Here is the description from the Web site:This applet is great for studying the volume of a rectangular prism and how to maximize it.   The graphic in the applet simulates a piece of paper that is 8 units by 10 units long. Drag your cursor in the top left quadrant. Dotted lines will show how much of the paper's corners will be cut.   Click the "Cut and Fold" button and the paper will fold into a box.

Try it: http://www.mste.uiuc.edu/users/carvell/3dbox/ Each time you click the “cut and fold” button, the applet folds up the box and tells you its volume and surface area.

By trial and error, use the applet to find the x that produces the largest volume.

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Making the box

You can get many sheets of cardboard and try many different x’s. You can cut and fold repeatedly and find the x by trial and error.

10 inches

8 inches

xx

You will need some tape. You will also need something to fill the box with. For example, you could use some sand or some granular cereal. For your first x – let’s say x = 0.5 inch – fill the folded box with the cereal. Then try a somewhat larger x – say 1.0 inches. Then pour the cereal from the first box into the second box and you will find (barring measurement and cutting/folding errors) that you will need to add more cereal. Then, repeat as necessary until you find that with larger x, you have too much cereal.

Surely there is a better way,

You can make a spreadsheet!

10 inches

8 inches

xx

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And so now we want to build a spreadsheet that multiplies L and W and H together for many successive values of x. We can start with a column for H, a column for W, a column for L, and a column for their product, the volume.

10 inches

8 inches

xx

The volume of a rectangular box depends on the length (L), width (W), and height (H).

It is easy to verify that the length of the rectangular base is (10 – 2x), the width is (8 – 2x), and the height is x. Therefore we can find the volume easily.

LW

H Volume = L × W × H

Making the spreadsheet: Getting started. What do we know?

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Question: What is the maximum volume of the box according to this chart? What if we used a smaller increment for x?

= cell with a number in it

= cell with a formula in it

10 inches

8 inches

xx

Maximize the Volume of a Box with L = (10 – 2x), W = (8 – 2x), and H = x, where L = length, W = Width, and H = height. V = L W H

Then the orange cells, where we want formulas to calculate the values. Start with the formulas in Row 4. In Cell C4, we want =10-2*B4 (note the “=“, which signals that the cell is a formula). In Cell D4, we want =8-2*B4. Complete the row. Then highlight the formula cells in that row, grab the handle in at the lower right corner, and drag down to Row 12 to complete Columns C, D, E and F.

Recreate this spreadsheet First the yellow cells. We place numbers in these cells.In Cell B4, we type 0 and in Cells B5 and B6, we type 0.5 and 1. Then we highlight these cells, grab the handle at the lower right corner and drag down to Row 12 to complete Column B.

B C D E F

2 X Value Length Width Height Volume of Box

3 inches inches inches inches cubic inches4 0.0 10.0 8.0 0.0 0.05 0.5 9.0 7.0 0.5 31.56 1.0 8.0 6.0 1.0 48.07 1.5 7.0 5.0 1.5 52.58 2.0 6.0 4.0 2.0 48.09 2.5 5.0 3.0 2.5 37.510 3.0 4.0 2.0 3.0 24.011 3.5 3.0 1.0 3.5 10.512 4.0 2.0 0.0 4.0 0.0

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10 inches

8 inches

xx

Highlight Cells B5 and C5 on Spreadsheet I, grab the handle at the lower right of corner of the block and drag it down to Row 12 to reproduce Spreadsheet II, which duplicates the volumes listed in the spreadsheet of Slide X.

Recreate these two spreadsheets

First make this adaptation on the spreadsheet of the previous slide. Note that Cell B5 is now a formula (orange). The formula adds the increment (0.5) to the previous value of x. So the formula is =0.5+B4. What formula goes into Cell C4. After entering the formula into Cell C4, test it by copying and pasting it into Cell C5.

I

II

Maximize the Volume of a Box with L = (10 – 2x), W = (8 – 2x), and H = x, where L = length, W = Width, and H = height. V = L W H

B C

2 X Value Volume of Box

3 inches cubic inches4 0.0 0.05 0.5 31.5

B C

2 X Value Volume of Box

3 inches cubic inches4 0.0 0.05 0.5 31.56 1.0 48.07 1.5 52.58 2.0 48.09 2.5 37.510 3.0 24.011 3.5 10.512 4.0 0.0

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10 inches

8 inches

xx

We can conclude that to the nearest 0.1 inch, the maximum volume (52.5 in3) occurs at x=1.5 in.

Now repeat the procedure using a smaller increment.

I

II

Maximize the Volume of a Box with L = (10 – 2x), W = (8 – 2x), and H = x, where L = length, W = Width, and H = height. V = L W H

B C

2 X Value Volume of Box

3 inches cubic inches4 0.0 0.05 0.1 7.6

B C

2 X Value Volume of Box

3 inches cubic inches4 0.0 0.05 0.1 7.66 0.2 14.67 0.3 20.98 0.4 26.59 0.5 31.5

10 0.6 35.911 0.7 39.712 0.8 43.013 0.9 45.814 1.0 48.015 1.1 49.816 1.2 51.117 1.3 51.918 1.4 52.419 1.5 52.520 1.6 52.221 1.7 51.622 1.8 50.723 1.9 49.524 2.0 48.025 2.1 46.326 2.2 44.427 2.3 42.228 2.4 39.929 2.5 37.530 2.6 34.931 2.7 32.332 2.8 29.633 2.9 26.834 3.0 24.035 3.1 21.236 3.2 18.437 3.3 15.738 3.4 13.139 3.5 10.540 3.6 8.141 3.7 5.842 3.8 3.643 3.9 1.744 4.0 0.0

10

10 inches

8 inches

xx

To reproduce the graph, highlight the values in Columns B and C, click on the Chart Wizard, and work through the prompts: select the XY scatter plot, and label the axes. After completing the prompts, clean up the presentation: right click on the axes, change the scale (maximum and minimum values and increment) and select an appropriate number of decimal places (under the tab labeled “number”).

Visualize the results by plotting them on a graph.

Maximize the Volume of a Box with L = (10 – 2x), W = (8 – 2x), and H = x, where L = length, W = Width, and H = height. V = L W H

B C

2 X Value Volume of Box

3 inches cubic inches4 0.0 0.05 0.1 7.66 0.2 14.67 0.3 20.98 0.4 26.59 0.5 31.5

10 0.6 35.911 0.7 39.712 0.8 43.013 0.9 45.814 1.0 48.015 1.1 49.816 1.2 51.117 1.3 51.918 1.4 52.419 1.5 52.520 1.6 52.221 1.7 51.622 1.8 50.723 1.9 49.524 2.0 48.025 2.1 46.326 2.2 44.427 2.3 42.228 2.4 39.929 2.5 37.530 2.6 34.931 2.7 32.332 2.8 29.633 2.9 26.834 3.0 24.035 3.1 21.236 3.2 18.437 3.3 15.738 3.4 13.139 3.5 10.540 3.6 8.141 3.7 5.842 3.8 3.643 3.9 1.744 4.0 0.0

0

10

20

30

40

50

60

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

length of cutout (x) inches

volu

me

of

bo

x, c

ub

ic i

nch

es

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Review of the First and Second Derivatives of a Function to find Local Maximum and Minimum

MOVING ON TO CALCULUS Review: Using first and second derivatives to find local maximums and

local minimums.

Example 1. Second Degree Polynomial:Let p(x)= –x2 + 4x + 5, Then p’(x) = –2x + 4, and p”(x) = –2 .

Set p’(x)= 0.Then –2x + 4 = 0, and x = 2.

Therefore the point (2, 9) is the local maximum, because p’(2) = 0 and p”(2) < 0.

Example 2. Third Degree Polynomial:Let p(x) = x3/3 – 5x2/2 + 6x + 1.Then p’(x) = x2 – 5x + 6, and p”(x) = 2x – 5.

Set p’(x) = 0.Then x1 = 2 and x2 = 3.

Therefore (for x1=2) the point (2, 5.67) is a local maximum, because p”(2) = − 1 < 0.

And (for x2=3), the point (3,5.5) is a local minimum, because p”(3) = 1 > 0.

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Use Calculus to Maximize the Volume of a Box with L = (10 – 2x), W = (8 – 2x), and H = x, where L = length, W = Width, and H = height.

Let’s return to the box.

The volume of the box as a function of the length of the cutout is:

V(x) = x (10 – 2x) (8 – 2x) = 4x3 – 36x2 + 80x = 4 (x3 – 9x2 + 20x)

Then the first derivative is:V’(x) = 4 (3x2 – 18x + 20)

The second derivative is:V”(x) = 4 (6x – 18)

Now let V’ = 0. We want to solve 3x2 – 18x + 20 = 0.

We recall the quadratic formula:Given the quadratic equation ,

the two roots are:

Problem: Build a spreadsheet to calculate x using the quadratic formula with

a = 3

b = −18

c = 20

a

acbbx

2

42

02 cbxax

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Use Calculus to Maximize the Volume of a Box with L = (10 – 2x), W = (8 – 2x), and H = x, where L = length, W = Width, and H = height.

Spreadsheet to solve 3x2 – 18x + 20 = 0, using

and V”(x) = 4 (6x – 18).

Recreate this spreadsheet

a

acbbx

2

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Part 1 finds the two roots of V’(x) = 0. Cell C6 uses the positive square root.Cell C7 uses the negative square root.

Part 2 determines whether the solution gives a local maximum or a local minimum. Cell B11 is an equation to repeat Cell C6; similarly Cell B12 calls Cell C7. Cells C11 and C12 use the equation for the second derivative.

Part 3 collects the results. Cell C15 calls the appropriate value for x (Cell C7). Cell C16 calculates the volume from Cell C15.

What do you think about Cell C6?

B C D2 Part 1. Finding x1 and x23 a 34 b -185 c 206 x1 4.52757 x2 1.472589 Part 2. Evaluating the second derivative

10 x first derivative11 4.5275 36.66 local min of V (x )12 1.4725 -36.66 local max of V (x )1314 Part 3. Calculating the maximum value15 x 1.4725 inches16 V (x ) 52.5138 cubic inches

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Use Calculus to Maximize the Volume of a Box with L = (10 – 2x), W = (8 – 2x), and H = x, where L = length, W = Width, and H = height.

Spreadsheet to solve 3x2 – 18x + 20 = 0, using trial and error

Recreate this spreadsheet

Part 1 finds a root. Cell C3 contains your guess for the root.Column D calculates the terms in the given quadratic equation, and Cell E7 sums them to find the value of the left side of the equation.Procedure: change Cell C3 until Cell E7 comes as close as you want to zero.

Part 2 calculates the volume of the box from the value of x determined in Cell E7.

You can find all the real roots of polynomial equation this way – quickly and easily with a spreadsheet.

What if you don’t remember the quadratic formula?

B C D E F

2 Part 1. Finding x for 3x 2-18x+20 = 03 x 1.4

4 3x 2 5.885 18x 25.26 20 20

7 3x 2-18x+20 0.68 inches8

9 Part 2. Volume from V(x) = 4x 3 - 36x 2 +80x

10 4x 3 10.976

11 36x 2 70.5612 80x 11213 V (x ) 52.416 cu. Inches

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End-of-Module Assignments

1. Why do the spreadsheets in Slides 7-10 use values of x in the range of 0 to 4.0? Discuss. Include in your discussion your thoughts about the question in Slide 13.

2. The spreadsheet in Slide 9 finds the value of x to the nearest 0.1 inch. Develop a spreadsheet that determines the value of x to the nearest 0.01 inch. Does your answer agree with the result determined by the spreadsheet in Slide 13?

3. A rectangular piece of cardboard measures 20 cm by 10 cm. Equal squares are cut out of each corner and folding up to make an open box. Find the maximum volume of the box.

4. We have a rectangular card of size a by b. Find the length of the square cutout that maximizes the volume of the associated open box.

5. Rectangular boxes come in a variety of shapes and sizes. Find a closed box or carton and determine its volume. Then unfold the box, and determine the size of the original rectangular piece of cardboard and how it was cut and folded to produce your box. Determine the maximum volume of a closed box that could be made from that piece of cardboard. Is it larger than the box that you started with? If so, why did the manufacturer make your box and not a larger one?