Among all right circular cones with a slant height of 15 , what are the dimensions (radius and height) that maximize the volume of the cone? The slant height of a cone is the distance from the oute edge of the base to the vertex Let V be the volume of the cone. What is the objective function in terms of the height of the cone, h ? \( \mathrm{V}=\frac{\pi}{3}\left(225 \mathrm{~h}-\mathrm{h}^{3}\right) \) (Type an expression Type an exact answer, using \( \pi \) as needed.) The interval of interest of the objective function is \( [0,225] \) (Simplify your answer Type your answer in interval notation.)
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To maximize the volume of the cone, we need to express the radius in terms of the height, h. Given the slant height \( s \) of 15, we can use the Pythagorean theorem: \( s^2 = r^2 + h^2 \). Substituting into the volume formula, we can express \( r \) as \( r = \sqrt{s^2 - h^2} = \sqrt{15^2 - h^2} = \sqrt{225 - h^2} \). This gives us the volume \( V(h) = \frac{\pi}{3}r^2h = \frac{\pi}{3}(225 - h^2)h = \frac{\pi}{3}(225h - h^3) \). The objective function in terms of the height h is: \[\mathrm{V}=\frac{\pi}{3}\left(225 \mathrm{h}-\mathrm{h}^{3}\right).\] The interval of interest for the height of the cone based on the slant height constraint is: \[ [0, 15\sqrt{2}] \text{ or } [0, 15] \text{ since, geometrically, h cannot exceed s.} \]
