The key thing is to see that the radius of 4 inches of the original circle shows up as the slant height of the cone.
let the height of the cone be h, and the radius of its base be r
r^2 + h^2 = 16, r^2 = 16-h^2
V = (1/3)πr^2 h
= (1/3)π(16-h^2)h
= 16πh/3 - πh^3/3
16πh/3 - πh^3/3 > 21
divide by -π/3
h^3 - 16h + 20.05 < 0
let's consider the equation
h^3 - 16h + 20.05 = 0
using my trusty cubic equation solver
http://www.1728.com/cubic.htm
I got h = -4.5, h = 3.08 and h = 1.44
looking at the graph of
h^3 - 16h + 20.05 < 0
I saw that it was negative for 1.44 < h < 3.08
so r has to be 2.55 < r < 3.73
Since the circumference of the base of the cone is the arc-length x as defined in your question
sub in the values of r into 2πr
if r = 2.55, x = 16.022
if r = 3.73, x = 23.44
check my arithmetic
Cone Problem Beginning with a circular piece of paper with a 4- inch radius, as shown in (a), cut out a sector with an arc of length x. Join the two radial edges of the remaining portion of the paper to form a cone with radius r and height h, as shown in (b). What length of arc will produce a cone with a volume greater than 21 in.3?
1 answer