Question

a hollow toy, with the dimensions shown in the figure, is to be stuffed with rigid foam. what is the maximum amount of foam that can be stuffed into the toy? round your answer to two decimal places. assume that both parts are stuffed and the foam does not expand or contract.

Explanation:

Step1: Identify the shapes and find the radius

The toy is composed of a cone and a hemisphere. The diameter of the base of the cone and the hemisphere is 4 inches, so the radius $r = \frac{4}{2}=2$ inches.

Step2: Calculate the volume of the cone

The height of the cone $h$ is $12 - 2=10$ inches. The volume formula for a cone is $V_{cone}=\frac{1}{3}\pi r^{2}h$. Substitute $r = 2$ and $h = 10$ into the formula: $V_{cone}=\frac{1}{3}\pi\times2^{2}\times10=\frac{40\pi}{3}$ cubic - inches.

Step3: Calculate the volume of the hemisphere

The volume formula for a hemisphere is $V_{hemisphere}=\frac{2}{3}\pi r^{3}$. Substitute $r = 2$ into the formula: $V_{hemisphere}=\frac{2}{3}\pi\times2^{3}=\frac{16\pi}{3}$ cubic - inches.

Step4: Calculate the total volume

The total volume $V$ of the toy (the amount of foam) is $V = V_{cone}+V_{hemisphere}=\frac{40\pi}{3}+\frac{16\pi}{3}=\frac{56\pi}{3}\approx58.64$ cubic - inches.

Answer:

$58.64$ cubic inches