We want to make a soda can out of aluminum in the shape of a cylinderThe top and bottom of the can will each be cut from congruent squares (same size and shape) as diagrammed belowThe side of the can will be formed from a rectangleWhat is the ratio of the height divided by the radius of the can that will minimize the total amount of aluminum used in making the can (including the waste in brown)? NOTE: The volume of the can should remain fixed regardless of the dimensions you use.The diagrams given don't matter
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If the volume of the can is fixed then the height is fixed relative to the radius i.eh V/(Pi.r^2) So the material needed is 2x(2r)^2 for the two squares + 2Pi.r.h for the rectangle which is 8r^2 + 2V/r which is independent of height for a fixed volumeNot reallyAnother way to ask the question is for a given surface area what is the maximum volume (for non-zero volume) As above S(fixed) 8r^2 + 2Pi.r.h so h (S-8r^2)/(Pi.r) Thus V Pi.r^2.h r(S-8r^2) when you sub in value of hwhich has a maximum for r^2 S/24 i.efor r^2 ( 8r^2 + 2Pi.r.h )/24 i.e when h 8r/Pi So min area for a given volume maximum volume for a given surface area is when h/r 8/Pi
