Investigation: The open box problem.
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
Past and Present ideas about Schizophrenia
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Geography Investigation: Residential Areas
Residential Areas How do Basingstoke's residential areas change, improve and reflect different urban models? Applied Understanding For my geography coursework I am going to study the different types of residential areas throughout Basingstoke; working from the Central Business District and following the main route through Basingstoke in a south westerly direction. When my coursework is complete I am hoping to come up with a conclusion about whether Basingstoke follows the concentric model or if it fits more to the criteria of the sector model. The central place theory is a geographical theory that tries to explain the size and spacing of humans, however, this theory only works when certain criteria are met, which in reality, aren't met, for instance: an isotropic, limitless amount of space, an evenly distributed population, evenly distributed resources, consumers all have same purchasing power and no provider of goods can earn excess profit. The concentric ring model (also known as the Burgess Model) is slightly more realistic than the central place theory and is actually based on a city, Chicago, Illinois. The theory puts forward that a settlement grows evenly (see Figure 1 below) , with the CBD in the middle, then light manufacturing, working class housing, middle class housing and then on the outskirts of the settlement, the high-class housing. At this stage I can
Compare the two poems 'Porphyria's lover' and 'My Last Duchess' by Robert Browning. In which way do they form part of a literary tradition?
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Persuasive essay
LW 26 Elias, Patrik 82 40 56 96 45 51 8 3 6 1 220 18.2 RW 89 Mogilny, Alexander 75 43 40 83 10 43 12 0 7 0 240 17.9 RW 17 Sykora, Petr 73 35 46 81 36 32 9 2 3 0 249 14.1 C 23 Gomez, Scott 76 14 49 63 1- 46 2 0 4 0 155 9.0 C 25 Arnott, Jason 54 21 34 55 23 75 8 0 3 2 138 15.2 LW 18 Brylin, Sergei 75 23 29 52 25 24 3 1 0 2 130 17.7 D 28 Rafalski, Brian 78 9 43 52 36 26 6 0 1 1 142 6.3 C 16 Holik, Bobby 80 15 35 50 19 97 3 0 3 0 206 7.3 RW 21 McKay, Randy 77 23 20 43 3 50 12 0 5 0 120 19.2 LW 11 Madden, John 80 23 15 38 24 12 0 3 4 1 163 14.1 D 27 Niedermayer, Scott 57 6 29 35 14 22 1 0 5 0 87 6.9 D 4 Stevens, Scott 81 9 22 31 40 71 3 0 2 0 171 5.3 C 12 Nemchinov, Sergei 65 8 22 30 11 16 1 0 2 0 70 11.4 RW 24 Stevenson, Turner 69 8 18 26 11 97 2 0 1 1 92 8.7 D 5 *White, Colin 82 1 19 20 32 155 0 0 1 0 114 0.9 D 6 O'Donnell, Sean ALL 80 4 13 17 0 161 1 0 2 0 67 6.0 MIN 63 4 12 16 2- 128 1 0 2 0 58 6.9 NJ 17 0
Open Box Problem.
Open Box Problem Aim During this project I will be 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. What is an Open Box 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. The card is then folded along the dotted lines to make the box. [image001.gif] [image002.gif] Names of things needed for investigation I will write-up my investigation in Microsoft Word and all formulae shall be calculated on Microsoft Excel and all table and graph will be produce in spreadsheets again in Microsoft Excel. Structure of investigation 1. Evidence: · Table · Graphs · Formulas 2. Evaluation 1. Evidence To obtain evidence I will be used a series of methods: · Table · Graphs · Formulae Part 1, Square I am going investigate 3 different sizes for the square open box. Once I have obtained all information on the 3 sizes I will look for patterns and try to formulate a rule to work out the largest volume for an open box square. The sizes that I will be using are: 1. 20 x 20 2. 40 x 40 3. 25 x 25 Because it is a square the length = width so, we can write this as L=1W therefore there is a ratio 1:1. I am going to begin by investigating a square with a side length of 20cm. Using
Fencing problem.
Problem specification A farmer has exactly 1000 metres of fencing. With this 1000 meter of fencing she wishes to fence off a plot of level land. She is not concerned with the shape of the plot, but must have a perimeter of 1000 m. her requirements are that the fence off the plot of land should contain the maximum area. Plan At the beginning of this experiment I shall begin experimenting with the simplest of all shapes. Triangles shall be investigated first. These triangles shall include equilateral triangle, an isosceles triangle, a right-hand triangle and a scalene triangle. I would find the area and plot the results upon a table. After investigation triangles I shall start with the regular shapes with four sides, a square, rectangles, a trapezium, a parallelogram and a rhombus. Once again areas of these shapes shall be found and recorded on a table. Now I shall continue experimenting different shapes by increasing the number of sides of each shape. These will include: triangles, quadrilaterals, a pentagon, a hexagon, a heptagon, and octagon, a nonagon and a decagon. I would take the whole investigation one step further and experiment with polygons shapes such as letters. For example a polygon shaped as an H, a polygon shaped as an E, and so on. These areas shall also be found and recorded. Finally I shall scrutinize a circle. The results of the area shall also be
Regeneration has had a positive impact on the Sutton Harbour area - its environment, residents and visitors. Discuss
Regeneration has had a positive impact on the Sutton Harbour area - its environment, residents and visitors I am investigating this topic because, as a class, we have studied the inner city areas of MEDC's. We have looked at some of the problems and considered how regeneration may help to solve some of them. Today, about 70% of the UK's population live in the city. This can lead to many problems, such as overcrowding, an increase in health and sanitation problems and much competition between businesses within a city and also between different cities that are trying to attract potential investors into the city. Trying to create a balance between the essential and desirable things in a city is hard to achieve when trying to keep both the residents and the city council happy. This is a problem that the Sutton Harbour area has had to try and overcome by using regeneration to make the place more attractive to businesses and tourists to help the council have a higher income, but also to suit the current residents. I will look at just how successful the regeneration of Sutton Harbour has been and find out the views of the people that have to live their, and why tourists were attracted to the area. Key Concepts Definitions Urban decay A region or building in a built up area that is of a poorer quality and condition to that of the surrounding area. Inner city The older
The Fencing Problem
Yr 10 GCSE Maths Coursework: The Fencing Problem In this investigation I plan to explore various polygons and deduce which type would yield the greatest area while complying with a perimeter of 1000m. I am only investigating regular polygons, because the otherwise would prove to be far too complex and involve an immeasurable amount of data. I will expand on this decision later in the investigation. I intend to find a general formula for the area of any polygon and prove it by applying it to each polygon I explore. The polygons I plan to investigate include: * Triangle (Isosceles, Scalene and Equilateral) * Rectangle * Parallelogram * Pentagon * Hexagon * Octagon * Decagon * A polygon with 15 sides (Pentadecagon) * A polygon with 20 sides (Icosagon) * A polygon with n number of sides (N-Gon) * Circle As you can see, I am working my way up chronologically in aspects of the number of sides on each polygon. For each one I will produce a table of data displaying the progressive areas for the shape; the areas will increase/decrease respectively according to the varying unit (e.g. 50) used for each instance of the shape. For every polygon I will produce a line graph presenting their progressive areas from the table of data. I will indicate the highest value on the graph (i.e. the highest area for the shape) and evaluate it accordingly. The area formulas for each
Beyond Pythagoras
Beyond Pythagoras This investigation is to study Pythagoras Theorem. I will try to find patterns and formulae to help predict Pythagorean Triples. About Pythagoras Pythagoras was a Greek Philosopher and Mathematician who is believed to have lived in the 6th century BC. He discovered many theorems but his most famous was: a2+b2= c2 What is a Pythagorean Triple? To answer this I first need to explain Pythagoras Theorem. Pythagoras States that in any right-angled triangle, a2+b2=c2. a is the shortest side, b the middle length side and c the hypotenuse (the longest side). A Pythagorean Triple is any set of integers that agrees this condition. For example 3, 4, 5 is a Pythagorean Triple because: 32+42=52 Because 32= 3x3= 9 42= 4x4= 16 52= 5x5= 25 9+16= 25 . The numbers 5, 12, 13 satisfy the condition: 52+122=132 Because 52= 5x5= 25 22= 12x12= 144 32= 13x13= 169 25+144= 169 The numbers 7, 24, 25 72+242= 252 Because 72= 7x7= 49 242= 24x24= 576 252= 25x25= 625 44+576= 625 2. (a) I found perimeter by using the formula: Perimeter= a+b+c P= 5+12+13 P= 30 Perimeter= a+b+c P= 7+24+25 P= 56 I found area using the formula: Area= (axb)?2 Area= 0.5x5x12 Area= 30 Area= (axb)?2 Area= 0.5x24x25 Area= 84 (b) I next put the results for perimeter and area into the table below. Length of Shortest Side (a) Length of Middle Side (b) Length of
