Anyone can help me in my additional mathematics 's project work?
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1)  the answer is h/r =2 or h = 2r\n\nThis is a classic optimization problem. You want the minimum surface area(SA) for a cylindrical shaped tin for a fixed Volume(V). This ratio of h/r that minimizes SA is independent of the volume though of course h and r are not independent of the volume.\n\n(1) SA = 2pi r^2 + 2pi r h\n\nV = pi r^2 h (for simplicity set V = 1)\n\nso (2) h = 1/(pi r^2)\n\nSubstituting this value for h into (1) gives:\n\nSA = 2pi r^2 + 2/r\n\nd/dr SA = 4pi r - 2/r^2\n\nd/dr SA = 0 when 2/r^2 = 4pi r\n\nthat is: (3) r = 1/(2 pi r^2)\n\nTo get the value of the ratio h/r divide equation (2) by equation (3), which yields a value of 2 to 1.\n\nYou are given the volume of the tins as 1 liter\n\n1000 = pi r^2 h,  but h=2r\n\n1000= 2 pi r^3 so r = (1000/2pi)^1/3 = 12.62 cm approx\nand h= 25.24 cm\n\n2) The answer is h/r = 8/pi\n\nIn this case your you have waste metal.\n\n2r is the diameter of the top of the tin. If the top and bottom are cut out of the square then the square will have to have an area of (2r)^2 = 4r^2. So the area A of material required for the tin is now: \n\n(1)   A = 2(4 r^2) +  2 pi r h  since the top and bottom are cut from squares.The procedure is the same in determining h/r, just a different function for the area of aluminum required\n\nAgain use V = 1 and so h = 1/(pi r^2)\n\nSubstituting this value of h in the above equation gives:\n\nA = 8r^2 + 2/r \n\ndA /dr = 16r -2/r^2\n\ndA/dr =0 when 16r = 2/r^2\n\nr^3 = 1/8 so r = 1/2 and h = 1/(pi r^2) = 4/pi\n\nSo h/r = 8/pi\n\nNow that you have the ratio h/r, you can determine the values of h and r when the volume of the can = 1000 cm^3\n\nNotice the value of h/r in this case is a little higher than in the no aluminum waste case. Of course, that makes sense\n\n3) This is also true for the equilateral triangle and hexagon\nYou need to determine the area of the equilateral triangle and hexagon having a maximum inscribed circle of radius r. The inscribed cirlcle will touch the midpoints of the edges of the regular polygon. We saw for the square this was 4 r^2. The general formula for the area of a regular polygon having N sides, in terms of r (its inradius, i.e. apothem) is:\n\nArea = r^2 x N x tan(pi / N)\n\nFor equilateral triangle N =3. For a hexagon, N=6. Plug these values in this formula and it will give you the area in terms of r When you calculate it, make sure your calculator is using radian measure, not degrees. So the area of aluminum required for the for the can if the top and bottom are cut from a regular polygon having N sides is\n\nA = 2 r^2 x N x tan(pi/N) + 2/r\n\nDetermine d A/dr. Set it equal to 0 and solve for r, and then solve for h.\n\nHere's a link about the Area of regular polygons.\n\nhttp://www.mathopenref.com/polygonregulararea.html
58231
http://www.studentquestions.com/homework_ans/4244.html
237126
http://mathforum.org/dr.math/\n\nThis is a good site where people ask questions and the questions are answered by volunteer professors.  May not have pictures but there are very good explanations.
1019971
Okay smarty here is a good site www.math2.org\n\ngood luck
792610
Try these;\nhttp://www.mathleague.com/help/geometry/geometry.htm\nhttp://www.coolmath.com/home.htm\nhttp://www.ics.uci.edu/~eppstein/geom.html
237248
How about doing your own homework ? ;)
53373
Here are a few sites you can check out:\nhttp://www.freemathhelp.com/\nhttp://www.mathgoodies.com/\n\nHope these sites can help you..
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Drop me your email. I will do one and send it to you. Easier than explaining...just look at my formulas...will only take a minute...really...\n\nCheck your email\n\nIs what I gave you what you want?
895105
kindly e-mail me the questions. I shall try to compile the result and give you in a day or so.
520168
Try this:\n\nhttp://www.algebra.com/
859713
You should go to www.AAAMath.com  good luck
578062
use the properties of equalities and inequalities.\n(subraction,additon,multiplication,division).
130573
Give me a better example and I may be able to help.:)
785120
http://www.freemathhelp.com/forum/viewforum.php?f=9\nhttp://www.purplemath.com/modules/
904464
Actually I would like to help but the location does matter...\nI cant come dre if its too far.\nAnyways ur idea is really nice....\nIf u r successful rply me...\nI would also try to open a math homewrok club.....\nSend me ur reply...ar\nruchir_rick_d@yahoo.com
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http://www.aaamath.com/ Try this. Hope it works!!
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try sosmath.com  they helped me before. if they don't have the answer you are looking for they have a list of links to tons of other math sites.  \n\nand there probably is someone on your campus that can help as well.  One on one help is always best...\n\nGood luck!