Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also find the maximum volume

Solution:

A sphere of fixed radius (R) is given.

Let r and h be the radius and the height of the cylinder respectively.

 

From the given figure, we have

h = 2 √ (R² - r²)

The volume (V) of the cylinder is given by,

V = π r²h

= 2π r² √ (R² - r²)

Therefore,

V = π r²h

= 2 π r² √ (R² - r²)

dV/dr = 4π r√ (R² - r²) + (2π r² (-2r)) / 2 √ (R² - r²)

= 4πr√ (R² - r²) - (2π r³) / √ (R² - r²​​​​​​​)

= [4π r √ (R² - r²​​​​​​​) - 2π r³] / √ (R² - r²​​​​​​​)

= (4π rR² - 6π r³) / √ (R² - r²​​​​​​​)

Now,

dV/dr = 0

⇒ (4π rR² - 6π r³) / √ (R² - r²​​​​​​​) = 0

4π rR² = 6π r³

r² = 2R²/3

Also,

d²V / dr² = {√ (R² - r²​​​​​​​) (4π R² - 18π r²) - (4π rR² - 6π r³) ((- 2R) / 2 √ (R² - r²​​​​​​​))} / (R² - r²)^3/2

Now, it can be observed that at r² = 2R²/3,

⇒ d²V / dr² < 0

Thus, the volume is the maximum when r² = 2R²/3

when, r² = 2R² / 3

Then, the height of the cylinder is

h = 2√ R² - 2R² / 3

= 2 √ (R² / 3)

= 2 R / √ 3

Hence, the volume of the cylinder is the maximum when the height of the cylinder is 2 R / √ 3.

Now finding the maximum volume of the cylinder:

V = π hR² - (π h³/4)

Replacing h with 2 R / √ 3, we get

V = (2π R³ / √ 3) - (π / 4 x 8R³/3√ 3)

V =  (2π R³ / √ 3) - (2 π R³/3√ 3)

V = (2π R³ / √ 3) × 2/3

V = (4π R³ / 3√ 3) 

Hence the maximum volume will be (4π R³ / 3√ 3) cubic units

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 17

Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. Also find the maximum volume

Summary:

Hence we have shown that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3. The maximum volume will be (4π R³ / 3√ 3) cubic units