
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?

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

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)=(1t^2)/(1+t^2)
Now volume is 1/3*pi*25*4*t^2*(5*(1+t^2)+5*(1t^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)^3u*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*(1t^2)/(1+t^2)=5*(1u)/(1+u) = 5*0.5/1.5 = 5/3
and the volume is 1/3*1000*pi*0.25/1.5^3

