Circular cone has slant height of 8 cm. What are dimensions of cone to product largest volume?

We know that the volume of the cone is given by
:

V = (1/3)*pi * r^2 * h such that r is the radius and h is
the height

But we know that the slant is given by s = sqrt(
h^2 + r^2)

==> s^2 = h^2 +
r^2

==> r^2 = s^2 -
h^2

Given that the slant s =
8

==> r^2 = 64- h^2

Now
we will substitute into the volume,

==> V =
(1/3)*pi* h * ( 64-h^2)

==> V = (1/3)pi*( 64h -
h^3)

==> (64pi/3)h - pi/3 *
h^3

Now we know that the maximum point if the derivatives
zero.

==> v' = 64pi/3 - pi*h^2 =
0

==> h^2 = 64pi/3pi =
64/3

==> h= 8/sqrt3 = 4.62
cm

Now we will find
r.

==> r^2 = s^2 - h^2 = 64 - 64/3 =
128/3

==> r= sqrt(128/3) = 8sqrt(2/3)= 6.532
cm

Then the dimensions of the cone are :
radius = 6.532 cm and the height h= 4.62 cm