Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #100

A rectangle of perimeter 36 cm is revolved about one of its sides. Find the rectangle dimensions and axis choice giving the greatest swept solid volume.

✅ Step-by-Step Solution

Final Answer: Radius side 12 cm; axis side 6 cm; Vmax=864π cm³.

Let the rotating distance (radius) be r>0 and the axis side be h>0. The perimeter condition gives r+h=18.
The swept solid is a cylinder with V=πr²h=πr²(18−r), 0<r<18.
V′=3πr(12−r)=0 gives the unique interior critical radius r=12.
V′ is positive before 12 and negative after 12, and limiting endpoint volumes vanish, proving the global maximum.
h=6 cm, so the rectangle is 12 cm×6 cm, rotated about its 6 cm side, and Vmax=π·144·6=864π cm³.

📋 CBSE Stepwise Marking Scheme

[1 mark] Let the rotating distance (radius) be r>0 and the axis side be h>0. The perimeter condition gives r+h=18.
[1 mark] The swept solid is a cylinder with V=πr²h=πr²(18−r), 0<r<18.
[1 mark] V′=3πr(12−r)=0 gives the unique interior critical radius r=12.
[1 mark] V′ is positive before 12 and negative after 12, and limiting endpoint volumes vanish, proving the global maximum.
[1 mark] h=6 cm, so the rectangle is 12 cm×6 cm, rotated about its 6 cm side, and Vmax=π·144·6=864π cm³.

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Class 12 Maths · Applications Of Derivatives · Very Long Answer #100

✅ Step-by-Step Solution

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Master Chapter 6: Applications Of Derivatives

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