Problem 6

A cone of revolution has an inscribed sphere tangent to its base and sloping surface. A cylinder is circumscribed about the sphere, with its base in the base of the cone. If the volumes of the cone and cylinder are

V_1

and

V_2

, respectively: (a) prove that

V_1\ne V_2

; (b) find the smallest possible value of

V_1/V_2

, and in this case construct the half-angle of the cone.

Step 3 of 4: Prove the lower bound and hence V1 ≠ V2

Previous step Next step

\dfrac{V_1}{V_2}\ge\dfrac43\iff (1+s)^3\ge8s(1-s^2)

Detailed analysis

Since the denominator is positive for $0<s<1$ , the inequality $V_1/V_2\ge4/3$ is equivalent to $(1+s)^3\ge8s(1-s^2)$ . Subtracting the right side gives $1-5s+3s^2+9s^3=(1-3s)^2(1+s)\ge0$ . Therefore $V_1/V_2\ge4/3>1$ , proving $V_1\ne V_2$ .