2017 AMC 12B Problems/Problem 14

Problem

An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?

$\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}$

Solution

Solution 1:

The top cone has radius 2 and height 4 so it has volume $\dfrac{1}{3} \pi (2)^2 \times 4$ .

The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so it has volume $\dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4$ .

Adding, we get $\dfrac{1}{3} \pi (16+32-4) = \boxed{ \textbf{E} \ \dfrac{44\pi}{3}}$ .

Solution by: SilverLion

Solution 2:

Find the area of the cone with the method in Solution 1. The area of the frustrum is $\pi\int_{0}^{4} \left(-\frac{1}{4}x+2\right)^2 dx=\frac{28\pi}{3}$

Adding, we get $\dfrac{44\pi}{3}$

Solution by IdentityChaos2020

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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All AMC 12 Problems and Solutions

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2017 AMC 12B Problems/Problem 14

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