The greatest volume of the cone made be 4372.7 m³.
Given, a right triangle whose hypotenuse is 22 m long is revolved about one of its legs to generate a right circular cone.
Let the base of the right triangle be, b
and the height of the right triangle be, h
Using Pythagoras Theorem, we get
22² = b² + h²
484 = b² + h²
Now, the height of the cone be, h and the radius of the cone be, b
Volume of the cone = 1/3πb²h
as, b² = 484 - h²
Volume = 1/3π(484 - h²)h
Volume = 1/3π(484h - h³)
On differentiating, we get
V' = 1/3π(484 - 3h²)
h² = 484/3
h = 22/√3
h = 22√3/3
b = 22√2/√3
Volume = 1/3π(22√2/√3)(22√2/√3)(22√3/3)
Volume = 1/9√3π(22³)(2)
Volume = 4372.7 m³
Hence, the greatest volume of the cone made be 4372.7 m³.
Learn more about Application of Derivatives here https://brainly.com/question/25120629
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