tag:blogger.com,1999:blog-15868947.post116691943198436177..comments2023-10-29T06:41:23.910-07:00Comments on Halfway There: Thinking inside the boxZenohttp://www.blogger.com/profile/09058127284297728552noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-15868947.post-1167055028693754142006-12-25T05:57:00.000-08:002006-12-25T05:57:00.000-08:00I was able to create a TI-83 program to solve the ...<I>I was able to create a TI-83 program to solve the general problem. </I><BR/><BR/>You are lucky that Mobius Stripper is not around, Jokermage.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-15868947.post-1166934597825825582006-12-23T20:29:00.000-08:002006-12-23T20:29:00.000-08:00I was able to create a TI-83 program to solve the ...I was able to create a TI-83 program to solve the general problem. It's a very simple problem if you can multiply polynomials, take simple derivatives and solve the quadratic formula.<BR/><BR/>Given a retangular paper with length L and width W...<BR/>The height of the box will be x.<BR/>The length of the box will be (L-2x)<BR/>The width of the box will be (W-2x)<BR/><BR/>Multiplying these together gives a volume equation of V = 4x^3 - 2(L+W)x^2 + (L*W)x<BR/>The derivative of this would be V' = 12x^2 - 4(L+W)x + (L*W)<BR/>Where the derivative is zero will be the optimal height.<BR/><BR/>x = (4(L+W) - Sqrt(16(L+W)^2 - 48(L*W))) / 24<BR/>Once you solve for x, plug it into the volume equation to calculate the max volume.<BR/><BR/>So did I get it right?Jokermagehttps://www.blogger.com/profile/02613078201972785157noreply@blogger.com