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The open box problem

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The open box problem An open box is to be made from a sheet of card. Identical squares are to be cut off at the corners so that the card can be folded into the open box. The diagram below shows the sheet of card and the four corners, which are to be cut off. There are two objectives I shall be investigating: 1. To determine the size of the four square cuts that will make the volume of the box as large as possible with any given square sheet of card. 2. To determine the size of the four square cuts that will make the volume of the box as large as possible with any given rectangular piece of card. First I shall investigate objective 1 as I think it will be easier to do. Objective 1: The square sheet of card. There are 2 ways to solve this problem: I can use algebra, or I can use trial and improvement. I think I will start by using trial and improvement; I will construct a series of tables and graphs and see if I can find any patterns. If so, then I will be able to come up with a hypothesis, which I can then test to see if I can solve this first objective. Then I will attempt to solve the same problem using algebra. Method 1: Trial and improvement To do this I need to make up some dimensions and then apply them to the square. I will choose the dimensions 6x6 to start with. I will call the length of the square cuttings x. The second drawing shows the open box with the dimensions included. x x 6cm 6-2x 6cm 6-2x x If we look at the second drawing we can see that to get the two lengths we have to take away 2x because of the two square cuttings on each side. ...read more.


V=L/6(L-L/3)(L-L/3) V=L/6(L^2-L^2/3-L^2+L^2/9) V=L/6(L^2-2L^2/3+L^2/9) V=L/6(L^2-5L^2/9) V=L/6(4L^2/9) V=2/27L^3 To check this is correct, lets try this out with the same square (6x6). So V=2/27L^3 16=2/27(6)^3 16=2/27(216) 16=16 So by two methods I have found the maximum volume for different squares, found the values of x and come up with an equation that works for any square. I will now move on to the rectangles problem. Objective 2: The rectangular piece of card Again there are two ways I can solve this problem, there is trial and improvement and the algebra way. I will start with the trial and improvement method, by drawing graphs and tables, and then spotting any patterns. If so, then I can come up with a hypothesis. Method 1: Trial and improvement The rectangular problem is a lot more complex then the square problem because the lengths and widths can be anything (not the same as in the square), so I think I will investigate this by choosing different ratios of length to width and see what I can come up with. I will start with a ratio of length to width 2:1 with dimensions of 6x3. x x 3 6-2x x 3-2x 6 So if I multiply the length by width by height I get the equation. V=x(L-2x)(W-2x). And if I replace the dimensions with L and W I get the equation V=x(6-2x)(3-2x). I will now create a table showing the different values of x and the volumes of this rectangle. X 0 0.5 1 1.5 V 0 5 4 0 We can see here that the maximum amount of x is between 0 and 1 so I will now draw another table showing that focuses in more on the values between 0 and 1. X 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 V 1.624 2.912 3.888 4.576 5 5.184 5.152 4.928 4.536 4 Here we can see that the maximum volume is at about 0.6 but it could go either way, so to get a better look at it I am going to draw a graph to show this. ...read more.


Method 2: Using Algebra The second method involves using algebra. To do this I will use the equation before and attempt to solve it. Somewhere in this equation I will need to use something called the gradient function. The gradient function is used to find the derivative in an equation. The gradient function is nx(n-1). I will give an example now as to how apply the gradient function in an equation. If I had the expression 4x^3, I would multiply the gradient (4) by the power (2) to get 8, then the power has 1 taken away from it. So 4x^3 would become 12x^2. I will now attempt to use algebra to find an equation for x. I already know the equation from before as V=x(L-2x)(W-2x) So V=x(L-2x)(W-2x) V= xLW-2x^2(L+W)+4x^3 Now I will apply the gradient function, nx(n-1). V=LW-4x(L+W)+12x^2 V=12x^2-4x(L+W)+LW Now you can see that a quadratic equation has formed, because there is a an expression with x^2, plus an expression with x, plus another expression with no x. V is replaced with 0 because when the volume is at it's max, the gradient (x) will be 0 (because the line will be at it's peak). The 0 is then replaced with x. I solve this quadratic equation using the formula -B+-sqrt(B^2-4AC) 2A A=12 B=-4(L+W) C=LW x=(4(L+W) +- sqrt((4(L+W)^2 - 48LW))/(24 24 x=(4(L+W) +- sqrt(16L^2 + 16W^2 - 16LW))/24 24 x=(L + W +- sqrt(L^2 + W^2 - LW))/6 6 x=L+W+-sqrt(L^2+W^2-LW) 6 The final equation uses the minus instead of a plus in the equation because if you used the minus the equation would not work. So x=L+W-sqrt(L^2+W^2-LW) 6 So I have now found an equation that satisfies all rectangles. So by using trial and improvement and algebra I have investigated the two problems involving squares and rectangles and have found ways to find out the maximum volume, and the value of x. ...read more.

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