In this problem we are asked to find the dimensions and volume of a box folded from an 8.5 x 14 piece of paper. We are then asked to find the largest volume by assigning a variable to one of the dimensions. In the following five part process, we will figure out how to solve these problems:
Part 1 - Area
To solve this problem we must first understand area. Area is the measure of the span of a two dimensional surface. For example, let’s say the surface we are working with is 3 inches wide and 4 inches long.
If we were to measure and draw lines both vertically and horizontally every one inch, we would have a grid with a series of 1in x 1in squares. The amount of squares in this grid would represent the area of the entire surface in inches2. If we counted the squares, there would be 12. So this rectangle has an area of 12 inches2.
To reduce the time of counting individual squares, we could say that we have 3 rows composed of 4 units. So we could add 4 units together 3 times (4units in row1 + 4units in row2 + 4units in row3)
or (4+4+4). 4+4+4 has a sum of 12, which is how many squares we counted originally.
To further reduce the time of adding, we can multiply. Since we are adding 4 units 3 times, we can say 4x3 = (4+4+4), or equally, 3x4 = (3+3+3+3). These problems also have a sum of 12.
In our case (rectangle), we can say that area is the number of rows multiplied by the number of columns, or width x length, or base x width. This formula will always equate to the number of squares on your grid, no matter the unit of measure (as long as they’re the same for base & width) and no matter the size of the rectangle or square.
Part 2 – Volume
Volume is area but with an added third dimension: height. In our example problem we had a 3x4 rectangle. If we added a third dimension, say 2 units high, our rectangle has now become a rectangular prism.
Since we are measuring a third dimension, our unit of measure must also gain a third dimension. Our 1in x 1in square will now become a 1in x 1in x 1in cube. Volume will measure the amount of these cubes that will fit into our rectangular prism (or box).
Since our box is 2 units high, we can find the volume by first finding the area (base x width = 12). 2 units high is like saying we are going to stack 2 boxes with an area of 12. We can add the two areas (base x width) + (base x width) = 24. If we count the number of 1in cubes that fit into our box, we would count 24.
If the height changes to 3, we would need to add the area 3 times: (base x width)+(base x width)+(base x width). Instead of adding all of these, we can use the shortcut of multiplication and say: base x width x 3. Since 3 is our height, we can adjust this formula for use with any box, cube or rectangular prism of any width, base or height as the formula: base x width x height = volume.
Base = 4, Width = 3, Height = 3 . Base x Width x Height = 4 x 3 x 3 = 36. A height of 3 = 36 cubes.
Part 3 – Finding measurements between the folds
In our problem, we are asked to find the dimensions and volume of a box folded from a 8.5 in x 14 in piece of paper. Using our formula for area (base x width), we know that the area of this piece of paper is 119. Base of 8.5 x Width of 14 = 8.5 x 14 = 119.
However, since we are folding the edges of this area to create height in the third dimension, our base area will decrease as our height increases. In our problem, the height is 3, so we will fold 3 inches from each edge to create height. (Highlighted yellow will become height, green is remaining base area).
This also means we need to remove each fold’s measurement from our base area (see above green area remaining after folds). In this case we will remove 3 inches from both sides of the 8.5 inch width. Since our paper is folded in half, the middle 3 inch fold will become the wall height between the base and the top, making the remaining (green) areas our base and top.
Now that we know where the folds are, we can find the area for our base and top. We can use our original measurements of 8.5 x 14 to find the new measurements of our base and top. Our base would be 8.5in minus the 3in folds from the bottom and top. If we add the fold measurements from bottom and top we get 6in, or 3in x 2. 8.5in – 3in – 3in = 8.5in – (2 x 3 in) = 8.5in – 6 in = 2.5in. So we can say our base has a width of 2.5in.
Likewise, we need to subtract the folds from the base area for the length. However, since we are folding the paper in half, our starting base length is 7in instead of 14in (14/2 = 7). Also since the center fold is the beginning of our wall, it is also the end of our base, meaning we only need to subtract the first 3in fold from our 7in length. 7in – 3in = 4in. So our base has a length of 4 in.
Now that we have the measurements for our cube (Base = 2.5in, Width = 4in, Height = 3in), we can find the area of our base using our formula (base x width). 4in base x 2.5in width = 10in2 (our answer is squared, because we are measuring how many 1in squares(2) fit into the area). So our base has an area of 10in2. If we find the measurements for our top (using the same methods described above), we will see that it also has an area of 10in2. Our problem gives us a height of 3in and our folded walls measure 3in. If we plug that 3in height into our (b x w x h) formula, for volume we get: 4in base x 2.5in width x 3in height = 30in3 (our answer is in inches cubed(3) because we are measuring how many 1in cubes fit into the box).
Part 4 – Using a variable and universal formula.
We now know that our box has a volume of 30in3. Is this the optimum volume? In order to find the answer to this, we must create a variable in our formula and reconfigure the formula to account for this variable. In this case our variable will be height. We can call our variable “X”.
We know that volume is base x width x height. If we add our variable for height, we can say our equation is base x width x (X).
We can also account for our folds changing the size of our base area. We know that width is originally 8.5in and that there are equal folds on both sides which will reduce the width of 8.5in. Since the folds will become height (our variable “X”) we can say: width = 8.5-2X.
So far, our equation for volume now looks like this: (base)(8.5-2X)(X), which still follows our original formula (base)(width)(height).
Our base of 7in (because the 14in paper is folded in half) also decreases with one variable (“X”) fold, so we could say: base = 7-X
Now our equation looks like this: (7-X)(8.5-2X)(X) = Volume. We can test this equation by solving for a volume we already know. In our original problem, the variable height (“X”) was 3in. If we replace all of the “X”s in our equation with 3s, we should come up with the same answer as before.
(7-3)(8.5-2x3)(3) = Volume
(4)(8.5-6)(3) = Volume
(4)(2.5)(3) = Volume
(10)(3) = Volume
30 = Volume = the same answer as our original problem
Part 5 – Applying the Formula
Now that we have our formula, (7-X)(8.5-2X)(X), we can begin our quest to find the largest volume. There are several methods to quickly resolving this issue. My preferred method is estimation mixed with logical elimination.
We have attempted solving the equation with 3 as our variable, so we can safely eliminate 3. But what if we tried 4 or 2? Let’s try both.
(7-4)(8.5-2x4)(4) = Volume
(3)(8.5-8)(4) = Volume
(3)(0.5)(4) = Volume
(12)(0.5) = Volume
6 = Volume
(7-2)(8.5-2x2)(2) = Volume
(5)(8.5-4)(2) = Volume
(5)(4.5)(2) = Volume
(10)(4.5) = Volume
45 = Volume
Our variable of 2 has resulted in a significantly higher volume. So let’s continue down that path and see what happens when our variable is 1:
(7-1)(8.5-2x1)(1)
(6)(8.5-2)(1)
(6)(6.5)(1)
(6)(6.5)
39 = Volume
Now we know that the variable 1 has a volume of 39, the variable 2 has a volume of 45, and the variable 3 has a volume of 30. Since 39 and 45 are both larger amounts than 30, we can logically estimate that the largest value will be somewhere between the variables 1 and 2. So let’s inspect this theory and split the difference to see what we get with the variable 1.5:
(7-1.5)(8.5-2x1.5)(1.5)
(5.5)(8.5-3)(1.5)
(5.5)(5.5)(1.5)
(30.25)(1.5)
45.375 = Volume
45.375 is greater than 45, so we know that the variable 1.5 results in a larger volume than both 1 and 2. Now we can try either side of 1.5, let’s try the variables 1.25 and 1.75:
(7-1.25)(8.5-2x1.5)(1.25)
(5.75)(8.5-2.5)(1.25)
(5.75)(6)(1.25)
(34.5)(1.25)
43.215 = Volume
(7-1.75)(8.5-2x1.75)(1.75)
(5.25)(8.5-3.5)(1.75)
(5.25)(5)(1.75)
(26.25)(1.75)
45.9375 = Volume
Since the variable 1.75 yields the largest volume of 45.9375in3, we can say that 1.75in is the optimum height for our box to a quarter of an inch.
If you feel a need for more accuracy, we can continue trying smaller and smaller fractions of an inch, until eventually we would reach a 1.71in variable which would yield a volume of 45.953172in3. 1.71in would be the optimum height value to 1/100th of an inch.
(7-1.71)(8.5-2x1.71)(1.71)
(5.29)(8.5-3.42)(1.71)
(5.29)(5.08)(1.71)
(26.8732)(1.71)
45.953172 = Volume
However, the variable can still be broken down further to 1/1000 of an inch, 1/10000 of an inch, and so on.
If we took all of our variables and plotted a graph for volume, we would see that the graph peaks at 45.953172 and drops on either side of that optimum volume.
Part 6 – Conclusion
Now that we have gathered all of our data, we can answer the original questions:
Question:
What are the dimensions of a 8.5 x 14 inch paper folded with 3 inch sides to create a box?
Answer:
· 8.5in – 6 in (2x 3in folds on top and bottom) = 2.5in. So we can say our base has a width of 2.5in.
· 7in – 3in (1x 3in fold on left) = 4in. So our base has a length of 4 in.
· And we are given the height of our folds: 3in
Our box has the dimensions 2.5in x 4in x 3in.
Question:
What would its volume be?
Answer:
· Volume = Base of 4 x Width of 2.5 x Height of 3 = 4x2.5x3 = 30
Our box has a volume of 30 inches3
Question:
Describe how to determine the dimensions of the box without actually building and measuring it.
Answer:
· Base = (7-3)
· Width = (7-2x3)
· Height = (3)
(7-3)(8.5-2x3)(3) = Volume (see Part 4 above for explanation)
Question:
Describe how you determined the volume of the box
Answer:
· See above answer and Part 4
Question:
Find the largest volume possible with a variable height
Answer:
· (7-1.75)(8.5-2x1.75)(1.75) = (5.25)(8.5-3.5)(1.75) = (5.25)(5)(1.75) = (26.25)(1.75) = 45.9375in3 = Volume. This is the largest volume using quarter-inch increments for the variable height.
· (7-1.71)(8.5-2x1.71)(1.71) = (5.29)(8.5-3.42)(1.71) = (5.29)(5.08)(1.71) = (26.8732)(1.71) = 45.953172 = Volume. This is the largest volume using 1/100th inch increments for the variable height.
· See Part 5.
All illustrations by Matthew Henderson using Adobe Illustrator. Source files available upon request.
All graphs by Matthew Henderson using Microsoft Office Excel – Original Graphs and Data Sheets attached.
Could i please get access to the graph images?
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