• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
• Level: GCSE
• Subject: Maths
• Word count: 1391

# Trays.The shopkeepers statement was that, When the area of the base is the same as the area of the four sides, the volume of the tray will be maximum.

Extracts from this document...

Introduction

Maths Coursework; Trays In this coursework candidates were given a task entitled "Trays." The task consisted of a shopkeeper's statement upon the volume of a tray which was to be made from an 18x18 piece of card. The shopkeeper's statement was that, "When the area of the base is the same as the area of the four sides, the volume of the tray will be maximum." By saying this, the shopkeeper basically meant that when the area of the base of the tray is equal to the total area of the sides the volume of the tray will be at its highest. We were told to investigate this claim. Plan. 1. I will investigate the different sizes of tray possible from an 18x18 piece of card. 2. After gaining my results I will then put them in a table. 3. I will try to spot any patterns from my table. 4. I will express any patterns or other formulae in mathematical notation. To investigate the different volumes given by different trays, I first decide to cut the corners in ascending order from 1-8. (The longest possible corner could only be 8 as after this there would be no base.) ...read more.

Middle

The answer was 256cm . The formula for this was (n - 2x) which out would be 18 (n) minus 2 times 1(x) squared. I than proceeded to work out the area of the sides, which would be essential in proving that the shopkeeper is right. To work out the are of the sides of the tray I used the formula 4x (n- 2x). Here again the "n" represents the size of card "18cm." The "x" represents the size of the corner. You have to times your answer by four as there are four sides. To work out the area of the sides for a corner sized 1x1cm the calculations would be: 4x (n - 2x) 4 x 1 (18 - 2 x 1) 4 ( 16 ) 64cm Corners Volume cm Area of base cm Area of sides cm 1x1 256 256 64 2x2 392 196 112 3x3 432 144 144 4x4 400 100 160 5x5 320 64 160 6x6 216 36 144 7x7 112 16 112 8x8 32 4 64 From my results I can see that in regards to the area of the base, the area lowers as the corner size is increased. ...read more.

Conclusion

However there is one main difference, the maximum volume is not given when both the areas of the base and area of sides is equal. Thus I graphed the area of the sides against the area of the base. You can see from my graph that the two area values crossed between 3 and 4 consequently the highest value lay between these two numbers if the shopkeeper was right. Corner Volume Area of base Area of Sides 3.05 589.2905 193.21 169.58 3.1 590.364 190.44 171.12 3.15 591.2235 187.69 172.62 3.2 591.872 184.96 174.08 3.25 592.3125 182.25 175.5 3.3 592.548 179.56 176.88 3.35 592.5815 176.89 178.22 3.4 592.416 174.24 179.52 3.45 592.0545 171.61 180.78 3.5 591.5 169 182 3.55 590.7555 166.41 183.18 3.6 589.824 163.84 184.32 3.65 588.7085 161.29 185.42 3.7 587.412 158.76 186.48 3.75 585.9375 156.25 187.5 3.8 584.288 153.76 188.48 3.85 582.4665 151.29 189.42 3.9 580.476 148.84 190.32 3.95 578.3195 146.41 191.18 4 576 144 192 4.1 570.884 139.24 193.52 4.15 568.0935 136.89 194.22 4.2 565.152 134.56 194.88 4.25 562.0625 132.25 195.5 4.3 558.828 129.96 196.08 4.35 555.4515 127.69 196.62 I conclude from my results that the shopkeeper's statement is not true on a 20x20cm card. ...read more.

The above preview is unformatted text

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

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related GCSE Open Box Problem essays

1. ## Maths Coursework

3 star(s)

Summary Table Size of square (x) Size of corner Proportion of square (x � size of corner) 12cm x 12cm 2 0.16~0.16 15cm x 15cm 2.5 0.16~0.16 18cm x 18cm 3 0.16~0.16 0.16 is 1/6. The size of the cut out square is approximately 1/6 of the whole square.

2. ## Maximum box investigation

The results are in the first table below. Length of the side of the corner square Length of the box Width of the box Height of the box Volume of the box (cm�) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

1. ## Open Box Problem

cm Width cm X cm Volume cm 0 10 40 0 0 0.5 9 39 0.5 175.5 1 8 38 1 304 1.5 7 37 1.5 388.5 2 6 36 2 432 2.1 5.8 35.8 2.1 436.044 2.2 5.6 35.6 2.2 438.592 2.3 5.4 35.4 2.3 439.668 2.31 5.38 35.38

2. ## Tbe Open Box Problem

Width (cm) Length (cm) Height (cm) Volume (cm�) 4.221 31.558 11.558 4.221 1539.598623 4.222 31.556 11.556 4.222 1539.599316 4.223 31.554 11.554 4.223 1539.59987 4.224 31.552 11.552 4.224 1539.600286 4.225 31.55 11.55 4.225 1539.600563 4.226 31.548 11.548 4.226 1539.600701 4.227 31.546 11.546 4.227 1539.6007 4.228 31.544 11.544 4.228 1539.600561 4.229 31.542 11.542 4.229 1539.600284 This shows that

1. ## THE OPEN BOX PROBLEM

Before investigating the rectangles in the ratio of 3:1 I need to prove through algebra how to get the maximum volume for the rectangle with a ratio of 2:1. I have already worked out that 1/4.73 of the width of the box gives me the maximum cut out size of

2. ## The open box problem

0.67 4.740652 0.68 2.64 2.64 0.68 4.739328 0.69 2.62 2.62 0.69 4.736436 0.70 2.6 2.6 0.7 4.732 0.71 2.58 2.58 0.71 4.726044 0.72 2.56 2.56 0.72 4.718592 0.73 2.54 2.54 0.73 4.709668 0.74 2.52 2.52 0.74 4.699296 0.75 2.5 2.5 0.75 4.6875 Size of cut-out Width Depth Height Volume 0.665

1. ## Investigate the volume of an open box constructed by one piece of rectangular card ...

length and another using this information to work out the maximum volume without constructing a net or working out the cut size. Rectangle card size I am now going to use a rectangle for the original card size; however there are an infinite number of combinations of width and length to try.

2. ## The Open Box Problem.

You can find a general formula allowing you to achieve maximum area by using algebraic expressions for the height length and width. 20 cm is the length and width of the square, therefore 20 = 2H + L Rearranging this it becomes L = 20 - 2H The formula for

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to