You can
put this solution on YOUR website!The formula
gives you the maximum volume for a cone cut of a sphere of radius
,
and for
, we get
.
The chemical filling up to the markings for 100-mL, 200-mL, 300-mL, and so on,
would be in the shape of similar cones, all scaled down versions of the big cone.
Their volumes (in cubic cm, which are equal to mL) would be directly proportional to the cube of the height in cm.
The function relating height to volume for the chemical would be
,
so if the height of the cone (in cm) is
,
we can calculate the heights (in cm) for the markings based on the proportion
-->
-->
That formula would work with heights or slant heights,
because, in similar cones, they would be proportional.
To get actual distances between markings on the outside of the cone,
we need to get the slant height of the cone.
There may be an easier way, but I will show how I calculated heights.
Here is a cross section of the sphere (with center O), and the cone (with its axis in green):
The radius is
, and
ia a fraction
of the radius.
In the right triangle ODB,
-->
allows to calculate
That is the radius of the base of the cone, squared.
The area of the base of the cone is
The height of the cone is
The volume of the cone
, is
Using calculus, we can figure out that the function
has a maximum
for
,
where
,
So
as stated in your posted question.
Most importantly, since
,
the height of the cone (in cm) is
.
The radius of the base of the cone, squared, is
The slant height,
can be calculated based on right triangle
.
-->
= about
Now we can substitute the values for
and
into
,
or using the slant height,
, of the cone,
to calculate the height,
, or the distance from the cone vertex,
for the markings for volumes
of 100 mL, 200 mL, and so on.
-->
and
-->
