Please help me with this problem?

Question

A right circular cone has an altitude of 12 ft and a base radius of 6 ft. A second cone is inscribed in the first, with its vertex at the center of the base of the first cone and its base parallel to the base of the first cone. Find the dimensions of the cone of maximum volume that can be so inscribed.

Answer 1

Ans.

altitude of inscribed cone = 4ft
radius of inscribed cone = 4ft

Soln:

let x = radius of base of inscribed cone
let y = altitude of inscribed cone

by ratio and proportion:

(12-y)/x = 12/6

or

y = 12 - 2x

Vol of cone = (1/3)* pi x^2 * y

= (pi/3)* x^2 * (12 -2x)

= (pi/3) * (12x^2 - 2x^3)

solve for dV/dx and equate to 0:

dV/dx = (pi/3) * (24x - 6x^2) = 0

solving for x:

24x - 6x^2 = 0
6x ( 4 - x ) = 0

x = 4 ft

y = 12 - 2x = 12 - 8 = 4 ft

Answer 2

Draw a right triangle representing one side of the first cone (height 12 ft, base 6 ft). If we draw a triangle inside this for the second code, with height h and radius r, we'll get a second triangle that splits the height into segments of length h and 12-h and splits the right angle, with a side of length r joining to the hypotenuse of the first triangle.

We now have two right-angled similar triangles, sharing a vertex (at the vertex of the first cone). One has sides 12 and 6, the other has corresponding sides 12-h and r. So we get
12-h = 2r, i.e. h = 12-2r.

V = 1/3 π r^2 h
= 1/3 π (12r^2 - 2r^3)
dV/dr = 1/3 π (24r - 6r^2)
= 0 if r = 0 or 4
d2V/dr^2 = 1/3 π (24 - 12r)
so volume is minimum at 0 (no surprise) and maximum at 4.
h = 12 - 2r = 4 also.
So the inscribed cone with maximum volume will have base radius 4ft and height 4ft.

Answer 3

The radius at the base is 6 and at the vertex is 0, so dr/dh = -h/2. The volume V of the second cone is V = π r² h / 3 = π (6 - h / 2)² h / 3 = π (36 - 6 h + h² / 4) h / 3 = π (12h - 2 h² + h³ / 12), so dV/dh = π (12 - 4 h + h² / 4) = 0 at the maximum, so solve the quadratic equation for h.

Since the guy above me did it a different way and came up with 4, I plugged 4 into the equation for dV/dh and--Voila!--I got zero, so his result is right, and apparently so is my method, despite the difference in calculation.