A cylinder is inscribed in a right circular cone of height 6 and radius (at the base) equal to 5. What are the dimensions of such a cylinder which has maximum volume?

1 Answer
May 5, 2015

The inscribed cylinder's volume will be maximized when its radius is r=10/3 units and its height is 2 units.

Let r be the radius of the cylinder inscribed in the cone and let y be the cylinder's height. The volume of the cylinder is,

V=B*h=(\pi r^2)*y

Place an edge of the cone at the origin and take the x to be the distance to the edge of the inscribed cylinder from the outside of the cone. The diagram below shows x.
enter image source here

From the diagram, the radius of the inscribed cylinder is r=5-x. We can also express the height of the cylinder, y, in terms of x because we know the slope of the slanted side of the cone. For every five units we move to the right we go up 6 units.

Slope= {rise}/{run}=6/5

\implies y=6/5x

Substitute r and y into the top expression for V

V=\pi(5-x)^2*6/5x=6/5pi(x^3-10x^2+25x)

To find the maximum value of V, find when d/{dx}V=0.

Keep in mind that 0\leq x \leq 5 so that the left edge of the inscribed cylinder stays properly within the cone.

d/{dx}V=6/5pi(3x^2-20x+25)=0

\implies3x^2-20x+25=0

Use the quadratic formula to find x

x={20\pm\sqrt{400-4*3*25}}/{2*3}={20\pm10}/6

The answer between 0 and 5 is x=5/3

You can verify that x=5/3 gives the point of a maximum by graphing V in terms of x.

At x=5/3,,,,, y=6/5x=2 and r=5-x=10/3