# Optimiztion=O

• Jul 30th 2009, 03:43 AM
User Name
Optimiztion=O
Hi to all! I've got quick but tricky question!
see this and then answer the following question if you could=)
http://i27.tinypic.com/seri9v.jpg
x is depth of cone!
how do we find volume of cone in terms of x? and maximum volume of cone?
and value of x which maximizes the volume of cone?
• Jul 30th 2009, 04:49 AM
songoku
From the picture, you can find the height of the cone. For the base radius, you can use pythagoras. (all in terms of x)

For maximum volume :
$\displaystyle \frac{dV}{dx} = 0$ -----> solve for x
• Jul 30th 2009, 05:51 AM
alunw
Attempted solution
Volume of right cone is 1/3*pi*r^2*h where r is radius of base and h is height.
Let theta be the angle between the axis of the cone and a line from the centre of the sphere to a point on the base circle. Then we have:

r=5*sin(theta)
x =5*cos(theta)
h = 5+x.

So volume is 1/3*pi*25*sin^2(theta)*(5+5*cos(theta)), where clearly theta is somewhere between 0 and pi/2 radians.

Put t = tan(0.5*theta) and use:

sin(theta)=2t/(1+t^2)
cos(theta)=(1-t^2)/(1+t^2)

Now volume is 1/3*pi*25*4*t^2*(5*(1+t^2)+5*(1-t^2))/(1+t^2)^3 =
= 1/3*pi*125*8*t^2/(1+t^2)^3

Now let us put u=t^2
Vol = 1/3*pi*1000*u/(1+u)^3

So differentiating wrt u we see that the maximum is at a root of

(1+u)^3-u*3*(1+u)^2=0
(1+u)^2((1+u)-3u)=0
u=-1 is impossible since u>=0
so u=1/2
but we had x = 5*cos(theta)=5*(1-t^2)/(1+t^2)=5*(1-u)/(1+u) = 5*0.5/1.5 = 5/3
and the volume is 1/3*1000*pi*0.25/1.5^3
• Jul 31st 2009, 04:22 PM
User Name
many thanks(Rofl)
• Jul 31st 2009, 08:30 PM
songoku
You're welcome ^^