Cone with
maximal volume

Task.

Design and make from poster board a cone with a
slant height S equal to 10 cm. The cone
should have the biggest volume possible.

Solution

The volume V of a cone with height h and radius
r is given by *V=1/*3*π*r ^{2}h.*

But
S^{2}= h^{2 }+ r^{2}, so r^{2 }= S^{2 }-
h^{2}

So,

We
want to maximize this volume.

Let's
use solver and nDeriv, to solve for h. (We now call it X.)

We
put in Y1:

\Y1=(π/3)*(100X-X^{3})

Then
in solver,

eqn:0=nDeriv(Y1,X,X)

We
have to make a guess for X (the cone's height). It will be less than 10.

Then
we press alpha solve

We
get X=5.7735026629… This is the height h of the cone.

What
next?

We
need to find the radius of the cone. We
use the Pythagorean theorem:

r^{2}
+ h^{2 }= S^{2 }

r^{2}
= S^{2 }- h^{2}

r = √
( S^{2 }- h^{2})

S =
10, so

r ≈
8.14619… cm

How
to design the cone?

A
cone is made from a sector of a circle, so we need to find how many degrees are
in the sector, namely, in the angle we will call A. Our circle will have a radius S of 10 cm,
which is the slant height of the cone.

Here
is the equation we need to solve:

angle
A/360°= (2πr)/(2π S)

So

angle A = (r/S)*360°.

Do you see it?

We
have r and S, so let's solve the equation for
angle A!

Now
we can build our cone!

What
is the volume of the cone?

Y1.

Calculus Index