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

Answer 1

Draw a sketch:

You can now make a cross section of the cone that shows the top of the cylinder:

where x is the radius of the cylinder and y is the height of the cylinder.

There are now similar triangle that you can set up a proportion with:

h/r = (h - y)/x

cross multiplying:

hx = r(h - y)

hx = rh - ry

ry = rh - hx

y = (h/r)(r - x)

The volume of the cylinder is π(x)²•(y) and substituting "(h/r)(r - x)" for "y" we get:

V(x) = π(x)²(h/r)(r - x)

V(x) = πhx² - π(h/r)x³

Take the derivative:

V'(x) = 2πhx - 3π(h/r)x²

Set the derivative equal to zero and solve:

2πhx - 3π(h/r)x² = 0

πhx(2 - (3/r)x) = 0

So x = 0 and x = 2r/3

Maximum volume when the radius of the cylinder is at (2/3)r

To find the height of the cylinder, plug in x = (2/3)r into y = (h/r)(r - x)