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^38¦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 20050402 10:12:17 