A sphere of radius r is centered at the origin. A right circular cone is inscribed in the sphere as shown in f

Question

http://i181.photobucket.com/albums/x85/ctti_3/untitled-33.jpg

Answer 1

Hi,

The maximum volume is answer d, 4000/3 π cu. in.

Let x = radius of the base of the cone. Then the area of its base is πx².

The height of the cone is 15 inches above the sphere's center plus v, the remainder of the height which can be found by looking at the right triangle formed by v,x, and r. Plugging in 15 for r, then v = √(15² - x²). So the total height of the cone is

h = 15 + √(225 - x²)

So the volume of the cone, Y, is found by the formula

Y1 = πx²(15 + √(225 - x²))/3

Use the MAXIMUM command under 2nd, CALC to find the solution.

This solves to a maximum point at x = 14.142131 where the Y value, the area of the cone, is 4188.7902.

If 4188.7902/π that gives 1333.33π which is 4000/3π.


I hope that helps!! :-)

Answer 2

Let
R = radius of sphere
r = base radius of inscribed cone
h = height of inscribed cone
V = volume of inscribed cone

We have

R = 15

The height of the cone can be divided into two sections. The top part is R, the radius of the sphere. The additional amount of height is y.

h = R + y

A right triangle can be drawn with sides

R = hypotenuse
r = horizontal leg
y = vertical leg

r² = R² - y²
_____________

The volume of the inscribed cone is:

V = (1/3)πr²h = (1/3)πr²(R + y) = (1/3)π(R² - y²)(R + y)

V = (1/3)π(R³ + R²y - Ry² - y³)

Take the derivative and set it equal to zero to find the critical points.

dV/dy = (1/3)π(0 + R² - 2Ry - 3y²) = 0
R² - 2Ry - 3y² = 0
3y² + 2Ry - R² = 0

(3y - R)(y + R) = 0
y = R/3, -R

But since y > 0, the negative solution is rejected.

y = R/3
h = R + y = 4R/3

r² = R² - y² = R² - (R/3)² = R² - R²/9 = 8R²/9

V = (1/3)πr²h = (1/3)π (8R²/9) (4R/3) = 32πR³/81

Plug in the value for R = 15.

V = 32πR³/81 = 32π(15³)/81 = 4000π/3

The answer is 4.
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