word problem using graphing calculator

Question

A cylinder is inscribed in a sphere of radius 1. Express the volume V of the cylinder as a function of the base radius r. State the domain. Using the graphing calculator, find the radius and height of the cylinder having the maximum volume. State the window used on the calculator to find this answer.

Answer 1

Hi Bianca:

 

to solve this problem, we need to calculate the volume of the inscribed cylinder.

 

Let us set

 

r = radius of cylinder

R = radius of sphere

h = height of cylinder

 

If we use spherical coordinates and call φ the angle measured from the vertical z axis, the edge of the top face of the cylinder is on the sphere, at a distance R from the center of the sphere (the origin O) and a horizontal distance r from the z-axis.  Therefore, using the right triangle:

 

sin(φ) = r / R

cos(φ) = h / (2R)

 

Now the volume of the cylinder is simply:

 

V = Πr²h = 2ΠR³ cos(φ) sin²(φ) = 2ΠR3 cos(φ) ( 1 - cos²(φ) )

 

which is only a function of the azimuthal angle φ.  When φ is zero, h is nearly R and r nearly zero, the cylinder turns into a very thin rod, and then a line.  As the angle φ increases, the cylinder gets shorter but thicker.

 

To find the maximum volume of the cylinder, you can use your calculator and vary the angle φ between 0 degrees and 90 degrees (when the cylinder turns into a very thin, flat pancake). 

 

Mathematically, we can calculate the derivative of the Volume as a function of angle, and equal the derivative to zero:

 

dV/dφ = 2ΠR3 sin(φ) ( 3 cos²(φ) - 1 )

The 2 solutions which zero out the derivative are φ = 0 (when the sinus is zero), and 3 cos²(φ) = 1

 

The first solution gives a zero volume (cylinder is a line), the second gives:

 

cos(φ) = ± 1 / √3, or

 

φ = 54.73 degrees and 125.26 degrees = 180o - 54.73o.  Same cylinder configuration.

 

The maximum volume is then:

 

V = 2ΠR³ cos(φ) ( 1 - cos²(φ) ) = 2ΠR³ * 2 / (3√3) = 2.4184 R³

 

You should approximate this maximum value when using your calculator.

 

Hope this helps!

 

Jean-Michel