There is a much easier approach to this problem than the one outlined in your steps
Clearly the radius of 4 inches of the original circle to be cut becomes the slant height of the cone.
then r^2 + h^2 = 16 ---> r^2 = 16-j^2
Volume = (1/3)pi(r^2)h
= (1/3)pi(16-h^2)h
= (1/3)pi(h) - (1/3)pi(h^3)
volume' = (1/3)pi - pi(h^2) = 0 for a max volume
then h = 1/√3 inches
and r^2 = 16 - 1/3
= 47/3
r = √(47/3)
so max volume = (1/3)pi(47/3)(1/√3) = 47√3pi/27 or appr. 9.472 cubic inches
back to the circle
let the central angle of the cut-out piece be α radians
then x = 4α
Circumference = 8pi
so the circumference of the cone = 8pi - x
but the circumference of the cone is 2pi(r)
so 2pi(r) = 8pi - x
r = (8pi - x)/(2pi)
a cone of height h and radius r is constructed from a flat, circular disk of radius 4 inches. by removing a sector AOC of arc length x inches. and then connecting the edges OA and OC. what are length x will produce the ocne of maximum volume, and what is that volume?
1. show that
r=(8pie-x)/2pie, h= radical 16-r^2,and
V(x)=pie/3((8pie -x)/(2pie))^2 *
radical (16-(8pie-x/2pie)^2)
2. shwo that the natural domain of V is 0<equal x <equal 16pie.
3. explain why the restriciton 0<equal x <equal 8pie makes sense in the problem situation.
4. use graphical methods to find where the cone has its maximum volume, and what that volume is.
5. confirm your findings in part 4 analytically. (Hint: use V(x)=(1/3)peir^2h, h^2+r^2=16, and the chain rule)
1 answer