Question

prove that the volume of greatest cone inscribe d in sphere is 8/27 times of sphere, where radius of sphere is R
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Answer 1

Answer:

Step-by-step explanation:

Let VAB be the cone of the greatest volume. Also it is clear for the maximum volume, the axis of cone must be along the diameter of the sphere.

Let OC = x.

Then

AC = √R² - x²

VC = VO + OC = R + x = Height of cone.

Let V be the volume of cone.

V = π/3(AC)²(VC) = π/3(R² - x²)(R + x)

differentiating the above function w.r.t x, we have

dV/dx = π/3[ (R² - x²) + (R+x) (-2x)]

=> dV/dx = π/3[R² - x² -2x(R + x)]

=> dV/dx = π/3[R² - 2Rx - 3x²]

For maximum or minimum, dV/dx = 0

=>   π/3[R² - 2Rx - 3x²] = 0

=> R² - 2Rx - 3x² = 0

=> (R - 3x)(R+x) = 0

=> R - 3x = 0  (∵ R + x ≠ 0)

=> x = R/3.

Let us compute d²V/dx at x = R/3

(d²V/dx) ₓ=π/3  = (-4/3)Rπ > 0

Therefore V is maximum at x = R/3.

Thus the maximum volume of cone = π/3(R² - x²)(R + x)

        = π/3(R² - R²/9)(R + R/3)

         = π/3(8R²/9)(4R/3)

          = 32πR³/81

          = 8/27(4/3πR³) = 8/27(Volume of Sphere).

Hence proved.