P.O.W. #4 Maximize Volume of Licorice Packaging                 James C. Gajniak

PAGE #1                                                                                 Math Teacher

P.O.W. #4, Page BP1-PP9                                                                  Instructor: Lisa Craig

 

                             Licorice Sticks and a Parabola

 

Problem:

 

A licorice candy manufacturer wishes to make Cardboard boxes which will hold the maximum amount of licorice sticks that are 12” long.

Process:

 

We will perform or address the following steps to determine the best sized 12” long box for holding the candy licorice.

 

  1. Make a four column table showing the height (h), the width (w), the l (length), and V (Volume) with the knowledge that Vol. = l*w*h;  that is, the Volume of the box = the length (always = 12”) * the width * the height.
  2. Using the values from the 4-column table, we will graph or plot V, Volume vs. h, height.
  3. By inspecting the graph in 2, we will determine the x-intercepts of V (h) and explain why those are the x-intercepts.
  4. By looking at the graph, we will estimate the greatest possible volume and specify the length, width and height for that volume.
  5. By looking at the table in Step #1, we will propose a

Function, w(h), connecting/relating the width and the height, w and h.

  1. Propose an equation for V (h) using the formula of V = l*w*h, where l = 12 always.

 

 

Each licorice stick, red and black, measures 12 inches long, and the cardboard stock measure 8” x 12”.   As their mathematician, I am to determine the dimensions of the container, the finished box, so that it will hold the maximum amount of licorice sticks.

 

The cardboard sheets look like this.
  1. Enter the data from Step #1 into the calculator and then graph V (h) equation from Step #6 above.
  2. Use the TRACE function on the calculator to find the maximum volume.
  3. Choose a larger WINDOW; and describe the general characteristics of the V (h) graph. Are there any other x-intercepts?  How many turns does the graph show?

 

P.O.W. #4 Maximize Volume of Licorice

 

 

 

 

 

 

 

We are to make two parallel folds where each fold line is 12 inches long, keeping in mind that we are forming an open box that will measure 12 inches long, and the width plus 2 times the height will always equal 8 inches. 

 

This is true because the cardboard stock measures 12” by 8”.

 

The volume of the open box is given by

 V = l * w * h, where V = volume, l=length=12, w=width, and h=height.

 

 

 

James C. Gajniak  Page #2.

Math Teacher

Instructor: Lisa Craig

 

Solution/Calculations:

 

2.     Graph of Volume vs. height using the values from Table of Possible boxes in Step #1.

 

Problem (concluded):

The cardboard sheet measures 8” by 12” so the height + width + height always will = 8” no matter how the box is folded.  The question is this:  How should the box be folded, that is, what should be the height, and the width, to hold the most candy.  Of course, the length will always be 12 inches to hold the 12” licorice sticks.  (Now, see Process, go to “upper right”.)

 

Page #3

James C. Gajniak

Licorice Stick Maximum Box Volume

 

Solution/Calculations:

 

1.     Table of Possible box sizes to hold the 12” licorice sticks

 

h (height)

 

0

½

1

2

2 ½

2 2/3

3

3 ½

3 ¾

4

 

 

W(width)

 

8=2*4

7

6=2*3

4=2*2

3

2 2/3

2=2*1

1

½

0

L (length)

 

12

12

12

12

12

12

12

12

12

12

 

Volume

=l*w*h

0

42

72

96

90

85 1/3

72

42

22 ½

0

 

 

 

 

 

 

Solution/Calculation (continued):

 

3.  The graph in Step #2 above shows a shape that appears to be a parabola; the x-intercepts are at (0,0) and (4,0). Since the h value is the first number of the coordinate pair, we have the left most intercept at h = 0, V, volume = 0.  By Algebra, when the height is 0, the Volume will also have to be 0 since

Volume = l*w*h = 12*w*0 = 0.  The second coordinate pair is (4,0).  Since 4 is the height and since h + h + w = 8 at all times, when we substitute, we have 4 + 4 + w = 8, or 8 +w=8.

The solution here is w=0.  Since Volume

James C. Gajniak, Page #4

is the product of l, w, and height, and since in this case, w=0, the Volume = 0.  (now, see step 4.,  above right)

 

 

 

6. Here is the equation I propose showing Volume as a “function of h”, V (h).  We know that Vol = l*w*h of the container. Since l=12 always, we have V = 12wh.  But from Step #5 above right, we have w (h)=2*(4-h), so substituting, we have

V (h) 12*w*h = 12*2*(4-h)*h = 24h(4-h) or

7. Entering the data from the chart, Step #1, into my calculator with h stored in L1 and

Volume (h), V (h) stored in L2, and setting Zoom to ZoomStat=9, I get a parabola pointing downwards, with the vertex at the high point (2,96) that is, h=2, Volume=96.

 

 

Reflection/Summary:

 

When we entered the data from the chart in Step #1, we saw the graph as in Step #2.  When we developed the

Volume (h), volume as a function of h in Step #5 (above

right), we had

When we graphed this equation, we got a smooth shaped parabola pointing downwards and this equation gave us the same graph as when we graphed the data points from the Chart in Step #1.

 

When we ran a quadratic regression analysis using our TI83 graphing calculator, we got

With a = -24, b=96, and c=0, we have

Solution/Calculation (continued):

 

  1. The graph in Step #2 above, shows the greatest volume at 96 cubic inches, with height=2 inches, width = 4 inches from the table, and length = 12 inches.  Since the graph is based on the table in Step #1, we can say that 96 cubic inches is shown in the table and graphed in Step #2.
  2. By inspection of the h vs. w columns in the data chart in Step #1, we see that:  w (h) = 2*(4-h)

      since w (1)=2(4-1)=6,

w (2)=2*(4-2)=4, w (3)=2*(4-3)=2, w (0)=2*(4-0)=8,

w (1/2) = 2*(4-1/2)=2*7/2=7, etc.and these calculations agree with the values of height and width in our table, Step #1..

 

 

 

 

 

 

  1. Using the calculator’s TRACE key with the equation for the volume, V (h), , we observe that the maximum volume occurs at height = 2 inches, and the maximum volume = 96 cubic inches.

 

9.     With a larger WINDOW, the general shape is that of an inverted parabola, with the vertex at (2,96) and the x intercepts at (0,0) and (4,0).  The graph shows only one turn and that is at the vertex.

 

 

 

 

Reflection/Summary(concluded):

 

In other words, the solution to maximize the volume of our cardboard is a parabola in the first quadrant denoted by,

Students with a TI 83, if entering the chart values into the computer and running a regression analysis will discover that the equation is as above, but by carefully setting up width, w, as a function of h, that is

w (h) = 2*(4-h) from Step #5 above, and then substituting into the volume equation, V=l*w*h, with l=12 will arrive at

 same equation,

 

This is a great problem to discover that the equation of a parabola uncovers not only the maximum volume possible, but all possible volumes depending on the value of the height.  Also, we discovered that the volume “depends on” the height.