Question

A cone is inscribed in a sphere with a diameter of 6. What is the maximum possible volume of the cone...
very hard problem

Answer 1

im in calc BC by the way
did you get 32pi/3?

Answer 2

See image.
basecircle of cone has radius 6sin(2 theta)
height of cone = 6 + 6cos(2 theta)

V(theta)=1/3 * pi * (6sin(2*theta))²(6+6cos(2 theta))

Now calculate the derivative and solve V'= 0...

Answer 3

Wait... its not necessarily an equilateral

Answer 4

In my drawing theta = alpha.

The base of the cone (DE) is not equal to DC or DE, but DC=DE, so u can use my formula to calculate the optimum volume.
Going to sleep now...Happy New Year! (already in Europe)

Answer 5

\[V(\theta)=\frac{ 1 }{ 3 }\pi r^2h=\frac{ 1 }{ 3 }\pi \cdot 36\sin^22\theta(6+6\cos2\theta)\]\[V(\theta)=72\pi \sin^22\theta(1+\cos2\theta)\]

Answer 6

maximum volume is \(\displaystyle \frac{32\pi}{3}\)

Answer 7

Derivative of V:\[V'(\theta)=72\pi2\sin2\theta \cdot 2\cos2\theta(1+\cos2\theta)+72\pi \sin^22\theta \cdot-2\sin2\theta=0\]
Divide everything by 144pi sin2θ: (because 2θ > 0, sin2θ >0, so dividing is OK)\[2\cos2\theta(1+\cos2\theta)-\sin^22\theta=0\]Use \[\cos^2x+\sin^2x=1\]to convert sin²2θ:\[2\cos2\theta+2\cos^22\theta+\cos^22\theta-1=0\]\[3\cos^22\theta+2\cos2\theta-1=0\]abc-formula:\[\cos2\theta=\frac{ -2 \pm \sqrt{4+12} }{ 6 }=-\frac{ 1 }{ 3 }\pm \frac{ 2 }{ 3 }\]so\[\cos2\theta=-1 \vee \cos2\theta=\frac{ 1 }{ 3 }\]Now, the first answer gives 2θ=π, which would mean a top angle op 180 degrees. This "cone" would have a volume of zero, so we have a minimum (not interesting).

Answer 8

The second answer, \[\cos 2\theta = \frac{ 1 }{ 3 }\]would mean\[\cos^22\theta=\frac{ 1 }{ 9 }\]and thus\[\sin^22\theta=\frac{ 8 }{ 9 }\]so\[\sin2\theta=\pm \frac{ 8 }{ 9 }=\pm \frac{ 2 }{ 3 }\sqrt{2}\]Of course, because\[0<2\theta<\pi\]after all, it has to be a cone, we see that \[\sin2\theta=\frac{ 2 }{ 3 }\sqrt{2}\]We can now use these values in the formula for the volume.
Maximum volume is:\[V=72\pi \frac{ 8 }{ 9 }(1+\frac{ 1 }{ 3 })=64\pi \frac{ 4 }{ 3 }=\frac{ 256\pi }{ 3 }\]
@sirm3d: my answer is different from yours!

Answer 9

Typo: \[\sin2\theta=\pm \sqrt{\frac{ 8 }{ 9 }}=\pm \frac{ 2 }{ 3 }\sqrt{2}\]