Maximum Volume: Making a Box from a Sheet of PaperDate: 10/21/1999 at 08:37:02 From: Chris Leahy Subject: Differentiation Hi, I'm working on a very important question that involves determining the largest possible volume when making a box out of a sheet of paper. To do this, you have to cut out squares in the corners of the paper. When using a square, I noticed that the side of the square equals a sixth of the size of the length of the paper. I only worked this out by noticing patterns in several tables that I drew up. I asked a friend of mine and apparently it has something to do with differentiation. I don't know anything about this, so could you explain it in simple terms to me so that I can explain how to show that the length of the side of the square equals a sixth of the length of a side of the piece of paper? I would be most grateful if you could help me with this. Kind regards, Chris Leahy Date: 10/21/1999 at 21:38:27 From: Doctor Jeremiah Subject: Re: Differentiation Differentiation is an algebra method to calculate the slope of a graph without graphing it. In your case we will say that the width of the piece of paper is W. I am also going to assume that we want an open box (no lid). The value you want to find is x. Then we get a diagram that looks like this: |<--------------W--------------> |<--x-->|<--A = W-2x-->|<--x-->| +-------+--------------+-------+ --------- | | | | | | | | | | | x | | | | | | +-------+--------------+-------+ | --- | | | | | | | | | | | | | | | | W B = W-2x | | | | | | | | | | | | +-------+--------------+-------+ | --- | | | | | | | | | | | x | | | | | | +-------+--------------+-------+ --------- The volume of the box is A*B*x: V = A*B*h = (W-2x)(W-2x)x = (W^2-4Wx+4x^2)x = W^2x-4Wx^2+4x^3 The graph of this equation will have a maximum. This is the value we are looking for. But we don't want to graph the equation and look for the maximum. Maximums and minimums are found when the slope is zero. When the slope is zero the graph is changing direction to go back the way it came and that is a maximum or minimum. If we differentiate this equation we get an equation of the graph's slope. And wherever that equation is equal to zero is a maximum or minimum. The differentiation operator looks like d/dx() so d/dx(V) means the differential of V. The differential of V is the slope of the graph of V. V = W^2x-4Wx^2+4x^3 d/dx(V) = d/dx(W^2x-4Wx^2+4x^3) = d/dx(W^2x) - d/dx(4Wx^2) + d/dx(4x^3) = W^2*d/dx(x) - 4W*d/dx(x^2) + 4*d/dx(x^3) = W^2*1 - 4W*2x + 4*3x^2 = W^2 - 8Wx + 12x^2 Remember that d/dx(V) is the slope and the maximum x is when the slope is 0, so: d/dx(V) = W^2-8Wx+12x^2 0 = W^2-8Wx+12x^2 Now we have to use the quadratic formula to find the values for the maximum and minimum values of x. max/min x = (-b +- sqrt(b^2-4ac))/(2a) = (8W +- sqrt((-8W)^2 - 4(12)(W^2)))/(2*12) = (8W +- sqrt(64W^2 - 48W^2))/24 = (8W +- sqrt(16W^2))/24 = (8W +- 4W)/24 = 12W/24 and 4W/24 = W/2 and W/6 The x = W/2 answer gives us: V = W^2x-4Wx^2+4x^3 = W^2(W/2) - 4W(W/2)^2 + 4(W/2)^3 = W^3/2 - 4W(W^2/4) + 4(W^3/8) = W^3/2 - 4W^3/4 + 4W^3/8 = W^3/2 - W^3 + W^3/2 = W^3 - W^3 = 0 So the x = W/2 answer is the minimum volume and the x = W/6 answer must be the maximum volume. V = W^2(W/6) - 4W(W/6)^2 + 4(W/6)^3 = W^3/6 - 4W(W^2/36) + 4(W^3/216) = W^3/6 - 4W^3/36 + 4W^3/216 = W^3/6 - W^3/9 + W^3/54 = 9W^3/54 - 6W^3/54 + W^3/54 = 4W^3/54 = 2W^3/27 The maximum volume (2W^3/27) of the box is found when the side of the box is W/6. If you need a better explanation, please write back. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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