The Open Box Problem
An open box is to be made from a sheet of card. Identical squares are
cut off the four corners of the card, as shown below.
[IMAGE]
The card is then folded along the dotted lines to make the box.
The main aim is to determine the size of the square cut which makes
the volume of the box as large as possible for any given rectangular
sheet of card, but first I am going to experiment with a square to
make it easier for me to investigate rectangles.
I am going to begin by investigating a square with a side length of 10
cm. Using this side length, the maximum whole number I can cut off
each corner is 4.9cm, as otherwise I would not have any box left.
I am going to begin by looking into going up in 0.1cm from 0cm being
the cut out of the box corners.
The formula that needs to be used to get the volume of a box is:
Volume = Length * Width * Height
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If I am to use a square of side length 10cm, then I can calculate the
side lengths minus the cut out squares using the following equation.
Volume = Length - (2 * Cut Out) * Width - (2 * Cut Out) * Height
Using a square, both the length & the width are equal. I am using a
length/width of 10cm. I am going to call the cut out "x." Therefore
the equation can be changed to:
Volume = 10 - (2x) * 10 - (2x) * x
If I were using a cut out of length 1cm, the equation for this would
be as follows:
Volume = 10 - (2 * 1) * 10 - *(2 * 1) * 1
So we can work out through this method that the volume of a box with
corners of 1cm² cut out would be:
(10 - 2) * (10 - 2) * 1
8 * 8 * 1
= 64cm³
I will the do the same for a 4x4x4 cube, 5x5x5 cube and finally a
BUT if we want the same perimeter (which we do) we have to take away a
Problem Statment:You have to figure out how many total various sized squares are in an 8 by 8 checkerboard. You also have to see if there is a pattern to help find the number of different sized squares in any size checkerboard.Process: You have to figure out how many total different sized squares you can make with a 8x8 checkerboard. I say that there would be 204 possible different size squares in an 8x8 checkerboard. I got that as my answer because if you mutiply the number of small checker boards inside the 8x8 and add them together, you get 204. You would do this math because if you find all of the possible outcomes in the 8x8, you would have to find the outcome for a 7x7, 6x6, 5x5, 3x3, and 2x2 and add the products of
I predict that in a two by two square the difference will always be 10
I shall prove his formula to be right by trying the next shape in the
PREDICTION: I think that the shape with the area of 6 cm2 will have 61
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The Golden Rectangle is a unique and important shape in mathematics. The Golden Rectangle appears in nature, music, and is often used in art and architecture. Some thing special about the golden rectangle is that the length to the width equals approximately 1.618……