Multiple choice

The height of the cone of maximum volume inscribed in a sphere of radius R is

  1. $\displaystyle \frac{\mathrm{R}}{3}$
  2. $\displaystyle \frac{2\mathrm{R}}{3}$
  3. $\displaystyle \frac{4\mathrm{R}}{3}$
  4. $\displaystyle \frac{4\mathrm{R}}{\sqrt{3}}$
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C Correct answer
Explanation

Volume V = 1/3 * pi * r^2 * h. In a sphere of radius R, r^2 = R^2 - (h-R)^2 = 2Rh - h^2. V = 1/3 * pi * (2Rh^2 - h^3). dV/dh = 1/3 * pi * (4Rh - 3h^2) = 0. h = 4R/3.

AI explanation

Using the relationship between the cone and sphere dimensions, the radius of the cone is r = √(2Rh - h^2), making the cone volume V = (1/3)π(2Rh - h^2)h = (1/3)π(2Rh^2 - h^3). Setting the derivative dV/dh = (1/3)π(4Rh - 3h^2) to zero gives 4R = 3h, so the height for maximum volume is 4R/3.

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