MEI Maths Item of the Month
February 2014
Cones from a Circle
An angle
θ is cut out of a circle of card to create two sectors: a major sector and a minor sector. The two sectors are then folded to make cones.
What angle
θ is required to obtain the largest value for the sum of the volumes of the two cones?
Solution
The first sector will have arc length r

. r

The resulting cone has radius
2π
and height is r
2  r
4π
2
 r
2π
4π
2   2
.
The cone has volume V
1
 r
3
24

 2
4π
2   2 .
The second cone has volume V
2
 r
3
24

(2π
  ) 2 4π 2  (2π   ) 2  r
3
24

(2π   ) 2 4π  2
Simplifying the sum of the volumes gives V
 r
3
24



2
4π
2   2  (2π   ) 2 4π 2
 d V

 r d 24
3





2
  2   2  x
3
4
 2  x
2

2( x
   x
 x
2 
( x
  3
4
 x
 x
2 



Solving d V d


0 numerically gives θ
= 2.036
4.247 being the maxima (in radians).
or θ
= π or θ
= 4.247
with θ
= 2.036
and θ
=
1 of 2 1 of 2 TB v1.0 01/05/14
© MEI

MEI Maths Item of the Month
Plotting the graph of V against θ :
This demonstrates the remarkable property that the sum of the volumes barely changes if the sector is between a quarter circle and three-quarters of a circle.
Over the range
 

2
to
 
3

2
the sum of the volumes varies by less than 0.8%.
2 of 2 2 of 2 TB v1.0 01/05/14
© MEI