Max/Min Problems with 3-D ShapesDate: 03/23/2009 at 13:30:57 From: Jesse Subject: Minimizing cylinder surface area I have a math project that I am doing under which I am trying to find the dimensions of a cylindrical can that will use the minimal surface area. The only time I have maximized area was with problems in which I had to determine the dimensions that would create the maximum area of rectangular space. In that case we used the perimeter as a constraint equation, solved for one of the variables and then substituted it into the area equation. In this case would the volume be my constraint equation? I know that volume of a cylinder is pi*r^2h and that the surface area is 2(pi*r^2)+(2pi*r*h). Do I solve for r or h? Date: 03/23/2009 at 15:05:53 From: Doctor Ian Subject: Re: Minimizing cylinder surface area Hi Jesse, You could choose either, and you'd get the same result, since either choice forces the other. It would be an interesting exercise for you to try using both h and r as the variable to substitute for. What do you notice? Presumably you're given the volume, since the point of a can is to hold a particular amount of some substance (beans, grease, cola, or whatever). Again, the idea is the same as for the 2-D problems you've solved in the past. Note that using numbers right away tends to make things more confusing. It's simpler if you just use variables everywhere. That is, if you have a cylinder, the volume and surface area are given by V = pi r^2 h and A = 2 pi r^2 + 2 pi r h = 2 pi (r^2 + r h) Suppose you're given a volume, and you want to minimize surface area. You can solve for height in terms of volume and radius, V = pi r^2 h V / (pi r^2) = h Now you can substitute into the equation for area: A = 2 pi (r^2 + r [V / (pi r^2)]) Now, note that you have two variables here: V and r. As soon as you specify a volume, you have only one variable, so you can apply the optimization techniques you've been learning. Nice, huh? On the other hand, suppose you're given a particular amount of surface area to use, and you want to maximize the volume. You might start by solving for height in terms of area and radius, A = 2 pi (r^2 + r h) . . [A / (2 pi) - r^2] / r = h Then you could substitute into the equation for volume: V = pi r^2 ([A / (2 pi) - r^2] / r) and you're in the same boat. You have two variables (A and r), but as soon as you specify an area, you have only one variable. The numbers just make it harder to see what's going on. If you leave everything unspecified until the end, you can just solve each kind of problem once, for all time. Does this make sense? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 03/23/2009 at 17:15:43 From: Jesse Subject: Thank you (Minimizing cylinder surface area) Thank you very much, this solved my issue completely. |
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