Question

Prove that the volume of the largest right circular cone that can be inscribed in a sphere of radius \(R\) is \(\dfrac{8}{27}\) of the volume of the sphere.

Hint:

Use r²=h(2R−h) for the inscribed cone, maximize V=(π/3)r²h with calculus, then divide by sphere volume.

Answer 1

Step 1: Understanding the Concept:
Set up the cone's height \(h\) and base radius \(r\) in terms of the sphere's radius \(R\) using the geometry of the inscribed cone, express the volume as a function of one variable, then maximize using calculus.

Step 2: Setting up the constraint:
For a cone inscribed in a sphere of radius \(R\) with its apex on the sphere and height \(h\) measured from the apex, the base circle's radius satisfies \(r^2 = h(2R-h)\) (from the right-triangle relation inside the sphere), where \(0<h<2R\).

Step 3: Writing the volume as a function of \(h\):
\[ V = \frac13\pi r^2h = \frac13\pi h(2R-h)h = \frac{\pi}{3}(2Rh^2-h^3) \]

Step 4: Differentiating and finding the critical point:
\[ \frac{dV}{dh} = \frac{\pi}{3}(4Rh-3h^2) = \frac{\pi h}{3}(4R-3h) \]
Setting \(\dfrac{dV}{dh}=0\) (and rejecting \(h=0\), which gives no cone):
\[ 4R-3h=0 \implies h = \frac{4R}{3} \]

Step 5: Confirming it is a maximum:
\(\dfrac{d^2V}{dh^2} = \dfrac{\pi}{3}(4R-6h)\); at \(h=4R/3\), this is \(\dfrac\pi3(4R-8R)=\dfrac\pi3(-4R)<0\), confirming a maximum.

Step 6: Computing the maximum volume:
At \(h=\dfrac{4R}{3}\): \(r^2 = \dfrac{4R}{3}\left(2R-\dfrac{4R}{3}\right) = \dfrac{4R}{3}\cdot\dfrac{2R}{3} = \dfrac{8R^2}{9}\).
\[ V_{\max} = \frac13\pi r^2h = \frac13\pi\cdot\frac{8R^2}{9}\cdot\frac{4R}{3} = \frac{32\pi R^3}{81} \]

Step 7: Comparing with the sphere's volume:
\[ \frac{V_{\max}}{V_{\text{sphere}}} = \frac{32\pi R^3/81}{(4/3)\pi R^3} = \frac{32}{81}\times\frac{3}{4} = \frac{96}{324} = \frac{8}{27} \]

Final Answer:
The largest inscribed cone has volume exactly \(\dfrac{8}{27}\) of the sphere's volume, as required. \[ \boxed{V_{\max} = \dfrac{8}{27}V_{\text{sphere}}} \]