Numerically Equal Volumes and Surface AreasDate: 06/04/2001 at 18:35:02 From: Irene Subject: Geometry Find all rectangular solids with integral dimensions, the volumes and surface areas of which are numerically equal. Date: 06/06/2001 at 02:29:20 From: Doctor Greenie Subject: Re: Geometry Hi, Irene - that was a curious problem and I had fun with it! Given a rectangular solid with length l, width w, and height h: volume = lwh and total surface area = 2lw+2lh+2wh You are looking for all integer triples (l,w,h) for which the volume and surface area are numerically equal. So you have lwh = 2lw+2lh+2wh Solving this equation for l in terms of w and h: lwh-2lw-2lh = 2wh l(wh-2w-2h) = 2wh and so we have 2wh l = -------- (1) wh-2w-2h In looking for integer triples (l,w,h) that satisfy this equation, we can assume that h <= w <= l, because a 3x4x5 rectangular solid is indistinguishable from a 5x3x4 or a 4x3x5 rectangular solid. And of course, h, w, and l are all positive integers. Since I have decided to call my shortest dimension h, I will try different integer values for h; for each choice of the value for h, I will look for integer values of w that are greater than or equal to h and for which equation (1) will give integer values of l greater than or equal to w. h=1: Equation (1) becomes l = 2w/(w-2w-2) = 2w/(-w-2) In this expression, the denominator is always negative; so there are no solutions with h=1. h=2: Equation (1) becomes l = 4w/(2w-2w-4) = 4w/(-4) In this expression the denominator is again always negative; so there are no solutions with h=2. h=3: Equation (1) becomes l = 6w/(3w-2w-6) = 6w/(w-6) In this expression, the denominator is positive for values w>6; so we can look for solutions with h=3 and w>6: h w l -------------------- 3 7 42/1 = 42 3 8 48/2 = 24 3 9 54/3 = 18 3 10 60/4 = 15 3 11 66/5 (not an integer) 3 12 72/6 = 12 For larger values of w, l would be less than w; so there are no further solutions with h=3. h=4: Equation (1) becomes l = 8w/(4w-2w-8) = 8w/(2w-8) In this expression, the denominator is positive for value w>4; so we can look for solutions with h=4 and w>4: h w l -------------------- 4 5 40/2 = 20 4 6 48/4 = 12 4 7 56/6 (not an integer) 4 8 64/8 = 8 For larger values of w, l would be less than w; so there are no further solutions with h=4. h=5: Equation (1) becomes l = 10w/(5w-2w-10) = 10w/(3w-10) In this expression, the denominator is positive for value w>=4; however, w>=h; so we can look for solutions with h=5 and w>=5: h w l -------------------- 5 5 50/5 = 10 5 6 60/8 (not an integer) 5 7 70/11 (not an integer) For larger values of w, l would be less than w; so there are no further solutions with h=5. h=6: Equation (1) becomes l = 12w/(6w-2w-12) = 12w/(4w-12) In this expression, the denominator is positive for value w>3; however, w>=h; so we can look for solutions with h=3 and (since w>=h) w >=6: h w l -------------------- 6 6 72/12 = 6 For larger values of w, l would be less than w; so there are no further solutions with h=6. There are no additional solutions with h>6. For h>6, and with w>=h, equation (1) gives values of l which are less than w; but our requirement is that l>=w. So we have the following set of solutions (h,w,l) to this problem; and this is the complete set of solutions: (3,7,42) (3,8,24) (3,9,18) (3,10,15) (3,12,12) (4,5,20) (4,6,12) (4,8,8) (5,5,10) (6,6,6) - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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