Out of a circular piece of paper, you wish to form a cone cup, so you cut out a circle wedge (with its extreme at the circle center) and join the resulting straight sides, forming a conical cup.

What size should the wedge be, to maximize the capacity of the cone?

Isn¡¯t this easiest to solve by making the volume of the cone a function of the angle and grinding through the differentiation?

<o:p> </o:p>

Let the radius of the wedge = slope length of the cone = 1 for simplicity¡¯s sake.

Calling the angle of the paper used (in radians), ¦È, we get that the circumference of the cone = arc length of the wedge used = ¦È.  Thus the radius of the base of the cone is ¦È/2¦Ð.

Pythagoras then tells us the height of the cone is ¡Ì(1-¦È^2/4¦Ð^2).

<o:p> </o:p>

Volume of cone, V= 1/3*¦Ð*r^2*h

                             = 1/3*¦Ð*¦È^2/4¦Ð^2* ¡Ì(1-¦È^2/4¦Ð^2).

                             = 1/12¦Ð(¦È^2 ¡Ì(1-¦È^2/4¦Ð^2).

<o:p> </o:p>

Max volume will be when dV/d¦È = 0

<o:p> </o:p>

dV/d¦È = 1/12¦Ð[2¦È* ¡Ì(1-¦È^2/4¦Ð^2) ¨C ¦È^3/(4¦Ð^2* ¡Ì(1-¦È^2/4¦Ð^2))]

<o:p> </o:p>

Setting this equal to zero gives 3¦È^3-8¦Ð^2*¦È = 0

Solving finds ¦È = 0 (minimum volume) or ¦È = 2¦Ð¡Ì(2/3) (maximum)

Thus the angle of the wedge to be removed

= 2¦Ð-¦È

= 2¦Ð(1-¡Ì(2/3)) rad

Posted by Alec on 2005-04-02 10:12:17