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Chapter 3: Problem 32

Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius \(r\).

Short Answer

Expert verified
The volume of the largest right circular cylinder that can be inscribed in a sphere of radius \(r\) is \(3\pi r^3/2\).

Step by step solution

01

Define the cylinder

Let's define the cylinder to be inscribed in the sphere as having a radius of \(a\) and a height of \(h\). The cylinder's center coincides with the sphere's, so the cylinder's top and bottom faces are tangent to the sphere. Therefore, the sphere's radius \(r\) forms the hypotenuse of a right triangle whose other two sides are \(a\) and \(h/2\). So \(r^2 = a^2 + (h/2)^2\). It can be rewritten as \(a= \sqrt{r^2 - h^2/4}\).
02

Express the cylinder's volume in terms of \(h\)

The volume \(V\) of a cylinder is given by \(V = \pi a^2 h\). Substituting \(a= \sqrt{r^2 - h^2/4}\) into the volume formula, it gives \(V = \pi (r^2 - h^2/4)h\). Further simplifying, we get \(V = \pi hr^2 - \pi h^3/4\).
03

Maximize the volume

To ascertain the maximum volume, differentiate \(V\) with respect to \(h\) and find the critical points by setting the derivative to 0. \(dV/dh = \pi r^2 - 3\pi h^2/4 = 0\). Solve for \(h\): \(h = \sqrt{4r^2/3}\). To determine that this gives maximum volume, consider the second derivative \(d^2V/dh^2 = -3\pi h/2\), which is less than 0 for positive \(h\), proving a maximum at \(h = \sqrt{4r^2/3}\).
04

Compute the maximum volume

Substitute \(h = \sqrt{4r^2/3}\) back into the volume equation \(V = \pi hr^2 - \pi h^3/4\) to get the maximum volume. This results in \(V_{\text{max}}= \pi (4r^3/3 * r^2 - 4r^3/3 * 4r^3/12) = 3\pi r^3/2\).

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