Asked by Lee Meyer
A cone is constructed by cutting a sector from a circular sheet of metal with radius 20cm. The cut sheet is then folded up and welded. Find the radius and height of the cone with the maximum volume that can be formed in this way.
Answers
Answered by
Steve
Given an arc of x radians, you should be able to convince yourself that you want the maximum volume
v = 1/3 pi r^2 h
where r = x/2pi * 20 = 10x/pi
and h = √(400-r^2)
so,
v = 1/3 pi (10x/pi)^2 √(400-(10x/pi)^2)
= 1/3 pi (10/pi)^3 x^2 √(4pi^2-x^2)
Note that as expected,
v(0) = 0 (zero radius base)
v(2pi) = 0 (zero height)
so, maximum volume must be somewhere in between.
dv/dx = 1000x/3pi^2 (8pi^2 - 3x^2)/√(4pi^2-x^2)
dv/dx = 0 when x=0 (minimum volume)
or x = √(8pi^2/3) (maximum volume)
so, for that value of x,
r = 10/pi √(8pi^2/3)
h = √(400 - 100pi^2 * 8pi^2/3) = 20/√3
max v is thus pi/3 (100/pi^2)(8pi^2/3) (20/√3) = 2000pi/(9√3)
As usual, check my math.
v = 1/3 pi r^2 h
where r = x/2pi * 20 = 10x/pi
and h = √(400-r^2)
so,
v = 1/3 pi (10x/pi)^2 √(400-(10x/pi)^2)
= 1/3 pi (10/pi)^3 x^2 √(4pi^2-x^2)
Note that as expected,
v(0) = 0 (zero radius base)
v(2pi) = 0 (zero height)
so, maximum volume must be somewhere in between.
dv/dx = 1000x/3pi^2 (8pi^2 - 3x^2)/√(4pi^2-x^2)
dv/dx = 0 when x=0 (minimum volume)
or x = √(8pi^2/3) (maximum volume)
so, for that value of x,
r = 10/pi √(8pi^2/3)
h = √(400 - 100pi^2 * 8pi^2/3) = 20/√3
max v is thus pi/3 (100/pi^2)(8pi^2/3) (20/√3) = 2000pi/(9√3)
As usual, check my math.
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