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# Investigation: The open box problem.

Extracts from this document...

Introduction

By Natasha Patel

Investigation: The open box problem

Problem:

An open box is to be made from a piece of card. Identical squares are to be cut off the four corners of the card to make the box. (As shown below)

Cut off

Fold lines

Aim:

Determine the size or the square cut which makes the volume of the box as large as possible for any given rectangular sheet of card.

Plan:

To start of with I will be using the trial and improvement method to experiment with different sizes of a square boxes. By doing this I will find out the size of cut off that will leave me with the largest volume inside the box. To find out the volume I will need to know the size of the cut off side and the base length.

x = length off the square cut off

L = original length off the square card

The formula that I will use to work out the volume is: Volume = (L-2X) ²X. The different sizes of cards that I will be using are 10cm, 11cm, 12cm, 13cm and 14cm. I will determine the size of x that will give the highest volume to 2d.p. After finding the highest value of X I will prove that my answer if right by using differentiation. Finally I will try and find a rule that allows me to find the highest value of X for a piece of square card and check that it works with any size of square card.

Trail and improvement

Size of card – 10cm by 10cm

X must be 0<X<5: This is because if X is 0 there would not be a side to fold and if X is 5 then there would be nothing left.

X = 1 X = 2 X = 3

V = (10-(2x1)) ²x1 V = (10-(2x2))

Middle

24 24

= 12L = 4L

24 24

X = L X = L

2 6

This cannot be the formula because L/2 = 0, there would be no card left to make the box with.

However this means that my formula I previously found was correct, X = L/6. So if I divide the length of any square by 6 it will give me the cut off that will leave the maximum volume.

Plan 2:

Now I will use trial and improvement to experiment with different sizes of a rectangular box. By doing this I will find out the size of cut off that will leave me with the largest volume inside the box. To find out the volume I will need to know the size of the cut off side and the width length.

x = length off the square cut off

L = original length off the width

The formula that I will use to work out the volume is:

Volume = (2L - 2X) (L - 2X) X for the 1:2 ratio

Volume = (3L - 2X) (L - 2X) X for the 1:3 ratio

Volume = (4L - 2X) (L - 2X) X for the 1:4 ratio

To find the different sizes of card I will use ratio. I will use three different sizes: 1:2, 1:3, 1:4. After this I will then look at the ratio, 1: n. I will determine the size of x that will give the highest volume to 2d.p. After finding the highest value of X I will prove that my answer if right by using differentiation. Finally I will try and find a rule that allows me to find the highest value of X for a piece of rectangular card and check that it works with any size of card.

Trail and improvement

Size of card – 1:2

Size of width 7

X must be 0<X<3.

Conclusion

Meanings of some of the mathematical words and symbols used in this investigation

Differentiation: This a method of calculations used to find the maximum or minimum of a point of a curve, also known as the gradient of a curve.

Trial and improvement: This is a method used to find a solution to a problem. It involves estimating number that might give the expected results and repeating this until the solution is found.

<: This is the sign for less than.

E.g. 2 < 4 is the same as saying 2 is less than 4.

>: This is the sign for more than.

E.g. 6 > 3 is the same as saying 6 is more than 3.

s.f.: This is short of significant figure. A significant figure is the first digit (*not zero) that you can reach, reading left to right. When approximating to a given number of significant figures, we use rounding off on the last significant number.

E.g. 2.3456 to 3s.f. = 2.35 however 0.067891 to 3s.f. = 0.0679.

d.p.: This is short for decimal place. A decimal place is the amount of numbers that come after the decimal point.

E.g. 2.3456 to 3.d.p = 2.346.

Gradient: The gradient of a line is a measurement of how ‘steep’ it is. The gradient of a line can be found by drawing right – angled triangles on the line.

Coefficient: Coefficient is the number that comes before a variable.

E.g. the coefficient of 5Y = 5.

Quadratic equation: A quadratic equation in x is an equation involving x to the power 2 but no higher powers of x. the general form of a quadratic equation is: ax² + bx + c = 0

E.g. 8x² - 4x + 6² = 0

When a quadratic equation has solutions and the equation is written in the general form, the solutions cab be found using the formula:

-b+/_√b 2 – 4ac

2a

Reciprocal: The reciprocal of a number x is the number 1/x.

E.g. the reciprocal of 3.4567 is 1/3.4567. The reciprocal of a number is the same as the multiplicative inverse of the number.

This student written piece of work is one of many that can be found in our GCSE Open Box Problem section.

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