Maximizing the Volume of a Box: Find Size of Square CutoutDate: 07/09/99 at 14:45:13 From: Anonymous Subject: Relation between maximum volume and length/width We are currently investigating the length of square cutout taken from each corner of a square/rectangle that gives a maximum volume for resultant box as the sides are folded up. I have found that for each rectangle with side ratios of 1:1, 1:2, 1:3, etc. there is a specific number that you divide the width of the rectangle by to give the length of the cutout that maximizes the box volume. 1:1 - 6 1:2 - approx. 4.7 1:3 - approx. 4.4 I tried to substitute 6 (for square) for x in the formula to give the volume of the box, i.e. x(l-2x)(l-2x). This rounded down to 2/27 (length cubed). I am now trying to find a general formula for the cutout that gives the maximum volume of the box for a rectangle of any width-to-length ratio. What do you suggest? I do not understand where the set denominator comes from. Do you know? Thank you for any help you can give. Anonymous Date: 07/09/99 at 18:20:12 From: Doctor Anthony Subject: Re: Relation between maximum volume and length/width Open Box -------- I have given a general solution to the problem with initial dimensions of the sheet a * b, and having found an expression for the volume in terms of x, the size of the cut, have differentiated to find the value of x that produces the box of maximum volume. If you are not familiar with calculus techniques, the value of x to give maximum volume can be found by plotting a graph of V against x and seeing where the volume is a maximum. If the card has dimensions a times b, and if x is the side of the square cut from each corner before folding up the sides, then the volume of the box is: V = x(a-2x)(b-2x) = x(ab - 2x(a+b) + 4x^2) = abx - 2(a+b)x^2 + 4x^3 To find a maximum point we can either plot a graph of V against x, or we get the maximum point by differentiating and equating to zero. dV/dx = ab - 4(a+b)x + 12x^2 = 0 You can use the quadratic formula to solve this equation 4(a+b) - sqrt(16(a+b)^2 - 48ab) x = --------------------------------- 24 (a+b) - sqrt((a+b)^2 - 3ab) x = ----------------------------- 6 (a+b) - sqrt(a^2 + 2ab + b^2 - 3ab) x = ------------------------------------- 6 (a+b) - sqrt(a^2 -ab + b^2) x = ----------------------------- 6 Even if you don't understand the above, use this formula (having put in the values of a and b) to see what is the best value of x. You could then plot the graph of V against x using a few values of x either side of this optimum value. You will find that the graph does peak at the value of x found by this formula. If a = b (square card) then: 2a - sqrt(a^2-a^2+a^2) 2a - sqrt(a^2) x = ---------------------- = -------------- 6 6 a side of square = --- = ---------------- 6 6 V = x(a-2x)(a-2x) = (a/6)(2a/3)(2a/3) = 2a^3/27 For rectangular cards we proceed as follows: Suppose for example you have a card of 18 * 12 inches, then a = 18, b = 12 and the value of x will be 30 - sqrt(324-216+144) x = ------------------------ 6 30 - sqrt(252) 30 - 6.sqrt(7) x = ---------------- = -------------- 6 6 x = 5 - sqrt(7) = 2.354 inches So you would cut a square of side 2.354 from each corner and then fold up the edges. This would provide a box of maximum volume. V = x(a-2x)(b-2x) = 2.354(18-4.708)(12-4.708) = 2.354 * 13.292 * 7.292 = 228.162 in^3 - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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