Question
A cylindrical can is to be made to hold 100 cubic inches. The material for its top and bottom costs twice as much per square inch as the material for the side. Find the radius and the height of the most economical can.
Answer
    This is an optimization problem where the cost of the can material must be minimized.  As with any problem of this type, you must first develop an expression for the quantity to be minimized, take the derivative with respect to the independent variable and set it equal to zero.<BR><BR>In this case, you know the volume must equal 100, so you can develop a relationship between the volume, the radius and the height of the can. Since ther volume is known, this is actually a relationship between the radius and the height of the can.  You can develop a second relationship between the radius and height of the can with the surface area.  You can then use these two relationships to create an equation for the cost of the can material using P for the cost per square inch for the sides and 2P for the cost per square inch for the ends.  This will give you a single realtinship between the cost of the material and either the radius or the height of the can, depending on which you eliminated with the first two relationships.<BR><BR>Relevant equations are:<BR>Volume = pi*r^2*h=100<BR>Sides surface area = 2*pi*r*h<BR>End surface area = 2*pi*r^2 (2 ends!)<BR><BR>Good luck!<BR>
Hello, <br /> <br />This is an additional answer provided by David Fulker: <br /> <br /> <br />Start by writing a formula for how many cubic inches the can holds. If you call its radius r and its height h, then your formula should have both h and r in it. It might be helpful to search for
