# Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere

**Solution:**

Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.

Let V be the volume of the cone.

Then, V = 1/3πr^{2}h

Height of the cone is given by,

h = R + AB

= R + √R² - r² [ABC is a right triangle]

Hence,

V = 1/3π r^{2}(R + √R² - r²)

= 1/3πr^{2} + 1/3π r^{2} (√R² - r²)

dV/dr = 2/3πrR + 2/3π r √R² - r² - 1/3π r^{2} - (- 2r)/√R² - r²

= 2/3πrR + (2πr (R² - r²) - πr^{3})/3√R² - r²

= 2/3πrR + (2πrR - 3πr^{3})/3√R² - r²

d^{2}V/dr^{2} = 2/3πR + [3√R² - r² (2π R^{2} - 9π r^{2}) - (2π rR^{2} - 3πr^{3}). (- 2r)/(6√R² - r²)] / 9(R² - r²)

= 2/3πR + [9(R^{2} - r^{2}) (2πrR^{2} - 9πr^{2}) + 2πr^{2}R^{2} + 3πr^{4}] / 27(R^{2} - r^{2})^{3/2}

Now,

dV/dr = 0

⇒ 2/3πrR + [2πrR^{2} - 3πr^{3}] / [3√R² - r²] = 0

⇒ 2/3π rR = [3πr^{3} - 2πrR^{2}] / [3√R² - r²]

⇒ 2R = (3r^{2} - 2R^{2}) / (√R² - r²)

⇒ 2R√R² - r² = 3r^{2} - 2R^{2}

⇒ 4R^{2} (R^{2} - r ^{2}) = (3r^{2} - 2R^{2})^{2}

⇒ 4R^{4} - 4R^{2}r^{2} = 9r^{4} + 4R^{4} - 12r ^{2}R^{2}

⇒ 9r^{4} = 8R^{2}r^{2}

⇒ r^{2} = 8/9 R^{2}

When, r^{2} = 8/9 R^{2}

Then,

dV/dr^{2} < 0

By second derivative test, the volume of the cone is the maximum when r^{2} = 8/9 R^{2}

When, r^{2} = 8/9 R^{2}

Then,

h = R + √R² - 8/9 R²

= R + √1/9 R²

= R + R/3

= 4/3 R

Therefore,

V = 1/3π (8/9 R^{2}) (4/3 R)

= 4/27 (4/3π R^{3})

= 8/27 (volume of sphere)

Hence, the volume of the largest cone that can be inscribed in the sphere is 8/27 the volume of the sphere

NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.5 Question 23

## Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

**Summary:**

Hence we have proved that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere

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