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 :
-----> solve for x
Hi to all! I've got quick but tricky question!
see this and then answer the following question if you could=)
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?
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