# Answer to Question #13357 in Calculus for Anandi

Question #13357

find the volume of the largest circular cone that can fit in a sphere of radius 3 cm.

Expert's answer

The volume of a right circular cone is V = 1/3 π r2 h. To apply the calculus you know you need to express this volume as a function of one variable. The right triangle ABC give the information you need.

Let

R = radius sphere

r = base radius cone

R + h = height cone

V = volume cone

_______

V = (1/3)πr²(R + h)

By the Pythagorean Theorem:

r² = R² - h²

Plug into the formula for volume.

V = (1/3)π(R² - h²)(R + h) = (1/3)π(R³ + R²h - Rh² - h³)

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

dV/dh = (1/3)π(R² - 2Rh - 3h²) = 0

R² - 2Rh - 3h² = 0

(R - 3h)(R + h) = 0

h = R/3, -R

But h must be positive so:

h = R/3

Calculate the second derivative to determine the nature of the critical points.

d²V/dh² = (π/3)(-2R - 6h) < 0

So this is a relative maximum which we wanted.

Solve for r².

r² = R² - h² = R² - (R/3)² = R²(1 - 1/9) = (8/9)R²

Calculate maximum volume.

V = (π/3)[(8/9)R²](R + R/3) = (8/27)πR³(4/3) = 32πR³/81

For R = 3 maximum volume is:

V = 32π(3³)/81 = 32π(27)/81 = 32π/3

Let

R = radius sphere

r = base radius cone

R + h = height cone

V = volume cone

_______

V = (1/3)πr²(R + h)

By the Pythagorean Theorem:

r² = R² - h²

Plug into the formula for volume.

V = (1/3)π(R² - h²)(R + h) = (1/3)π(R³ + R²h - Rh² - h³)

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

dV/dh = (1/3)π(R² - 2Rh - 3h²) = 0

R² - 2Rh - 3h² = 0

(R - 3h)(R + h) = 0

h = R/3, -R

But h must be positive so:

h = R/3

Calculate the second derivative to determine the nature of the critical points.

d²V/dh² = (π/3)(-2R - 6h) < 0

So this is a relative maximum which we wanted.

Solve for r².

r² = R² - h² = R² - (R/3)² = R²(1 - 1/9) = (8/9)R²

Calculate maximum volume.

V = (π/3)[(8/9)R²](R + R/3) = (8/27)πR³(4/3) = 32πR³/81

For R = 3 maximum volume is:

V = 32π(3³)/81 = 32π(27)/81 = 32π/3

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