Open Box ProblemDate: 06/22/2003 at 16:46:14 From: Georgia Wright Subject: Open box problem I am not sure how to find the formula for the greatest volume box you can make from a sheet of cardboard with different-sized corners cut out of it. You need calculus to solve it but I am only 13 so I need some help. i have figured that for a sheet of card that iis n by n that the formula is 2times n^3 all divided by 27 i found this out simply by finding the biggest cmsquared you need to cut out of a 20 by 20 square (which is 3.33333) and simplifying (L-2L/6)(L-2L/6)L/6 with L being the length of the sheet of card, i know there is an easier way to solve it, also a way to solve it for a rectangular sheet of card, i have read about calculas at your site, it is a bit hard to follow , and i am not sure how to apply this to my problem, i would be very grateful if you could help me it will help me get my head round harder maths.Thankyou for your time Date: 06/22/2003 at 21:05:38 From: Doctor Ian Subject: Re: Open box problem Hi Georgia, You don't actually _need_ calculus to solve a problem like this. It just makes the solution easier to find. Suppose we have a card that is L by W, and we want to cut some square corners (of length x by x) out so we can fold it into a box: +-------------+ --- | | | +--+ +--+ | | | | W +--+ +--+ | | | x | +-------------+ --- |--------L----------| The volume of the box will be volume = height * area of base = x * ((L - 2x) * (W - 2x)) Now, if you choose particular values for L and W (e.g., L=10 and W=6), then you get an equation that gives the volume as a function of x: volume = x * (10 - 2x) * (6 - 2x) Suppose we plot volume for various values of x? What will the graph look like? For one thing, we know that the volume will be zero when x=0. Do you see why? It will also be zero when x=5, and when x=3. So we can plot those points right away: | - | - | - | - | - | --*--|--|--*--|--*--|--|--|--|--|--|--|-- Of course, x can't get all the way to 5. We can't make x larger than half the length of the shorter side of the box. And we can't make x negative. So we really only care about values of x between x = 0 and x = 3: | - | - | - | - | - | --*--|--|--* So, what happens in between? Let's try a couple of values. When x = 1, we get volume = 1 * (10 - 2*1) * (6 - 2*1) = 1 * 8 * 4 = 32 And when x = 2, we get volume = 2 * (10 - 2*2) * (6 - 2*2) = 2 * 6 * 2 = 24 Let's plot those two points: | 50 - | 40 - | * 30 - | * 20 - | 10 - | --*--|--|--* 1 2 3 How about right in the middle, at x = 1.5? volume = 1.5 * (10 - 2*1.5) * (6 - 2*1.5) = 1.5 * 7 * 3 = 31.5 | 50 - | 40 - | * 30 - * | * 20 - | 10 - | --*---|---|---* 1 2 3 How about closer to the left side of the range, at x = 0.5? volume = 0.5 * (10 - 2*0.5) * (6 - 2*0.5) = 0.5 * 9 * 5 = 22.5 | 50 - | 40 - | * 30 - * | * * 20 - | 10 - | --*---|---|---* 1 2 3 Now, if you keep plotting points, you'll find that you end up with a curve that looks kind of like a distorted parabola. (If it were actually a parabola, we would know from symmetry that the highest point would be halfway between x = 0 and x = 3, at x = 1.5. Clearly that isn't the case here.) The point of all this is that the highest point on this curve will occur at the value of x that gives you the greatest volume for given values of L and W. Does that make sense? So you can solve the problem by completing the graph, without using calculus. Just to give you a glimpse ahead, at any point P on a curve, we could choose two nearby points A and B (also on the curve) and compute the slope of the line segment between them, as shown below: A . . *. change in y P . change in y slope = ----------- * . change in x B....... change in x For most curves that we deal with, as the points A and B get closer to P, the slope of the segment AB approaches the slope of the curve itself at P. And here's a very useful fact: at the very top of a curve like our distorted parabola, this slope will be horizontal, that is, equal to zero: P A.....B * * * * And this is where calculus comes into it. If we compute the 'derivative' of the function v(x) = x * (10 - 2x) * (6 - 2x) = x * (60 - 12x - 20x + 4x^2) = x * (4x^2 - 32x + 60) = 4x^3 - 32x^2 + 60x we get 3-1 2-1 1-1 dv/dx = 3*4x - 2*32x + 1*60x = 12x^2 - 64x + 60 This equation tells us the _slope_ of the curve at any point. If we'd like to know the values of x where the slope is horizontal - that is, where the curve reaches a maximum (or minimum) - we can set the derivative equal to zero and solve for x. 0 = 12x^2 - 64x + 60 0 = 3x^2 - 16x + 15 Using the quadratic formula, -(-16) +/- sqrt((-16)^2 - 4(3)(15)) x = ----------------------------------- 2(3) 16 +/- sqrt(256 - 180) x = ---------------------- 6 16 +/- sqrt(76) x = --------------- 6 16 +/- 8.7 x = ---------- 6 = 24.7/6 or 7.3/6 The first of these is out of range (i.e., it's greater than half the width of the box). So the second must be the value of x that we're looking for. Substituting it back into our equation, volume = 1.2 * (10 - 2*1.2) * (6 - 2*1.2) = 1.2 * 7.6 * 3.6 = 32.8 which is higher than any of the other values we've tried. In fact, this is the same point you'd find by completing the graph. But as you can see, this is a lot quicker, and this illustrates one of the most important uses of differential calculus. That is, if you can find a way to express one quantity as a function of another, you can use derivatives to quickly find the largest or smallest possible values of the function. Since we're often interested in maximizing quantities (like the amount of money that we can charge for something) or minimizing quantities (like the amount of money that we'll have to pay for something), these kinds of problems are called 'optimization' problems. In the case of the box, we're trying to find the 'optimal' value of x, i.e., the one that gives us the greatest volume. Note that we picked specific values for L and W. But you said you were looking for a formula that would cover any case. Can we find one? Without using calculus, it would be awfully hard. But using calculus, we can do this: 1) Expand the formula to get a polynomial. volume = x * (L - 2x) * (W - 2x) = x * (LW - 2Wx - 2Lx + 4x^2) = x * (4x^2 - 2(L+W)x + LW) = 4x^3 - 2(L+W)x^2 + LWx 2) Find the derivative and set it equal to zero. d volume -------- = 12x^2 - 4(L+W)x + LW dx 0 = 12x^2 - 4(L+W)x + LW 3) Use the quadratic formula. -[-4(L+W)] +/- sqrt([-4(L+W)]^2 - 4[12][LW]) x = -------------------------------------------- 2[12] I'll leave it to you to simplify this. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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