What reason do you have to think that the height of the cone must be less than or equal to the radius of its base? Of course, h cannot be larger than the diameter of the sphere but that is '2', not '2r'.
Couldn't really put the constraints on the formula for surface area of a cone.If you fit the cone with the largest possible surface area (lateral area plus area of base) into a sphere, what percent of the volume of the sphere is occupied by the cone
So:
$V(sphere) = \dfrac{4}{3}\pi r^3$
I guess I could just take $r = 1$ then it is: $V(sphere) = \dfrac{4}{3} \pi$
$ A(cone)= \pi r (r + \sqrt{h^2 + r^2}) $
And here we have to define the domain, as I understand; so $ r \leq 1$, $h \leq 2r$
Is this correct thinking?
Yes, you're right.
Well, let's say then:
$V(sphere) = \dfrac{4}{3}\pi R^3$
$R = 1$
$V(sphere) = \dfrac{4}{3} \pi$
$ A(cone)= \pi r (r + \sqrt{h^2 + r^2}) $
Domain: $ r \leq 1$, $h \leq 2r$ or $ h \leq 2 $
HallsofIvy, could you help me go from here? I'm not sure what to do next. I have two variables in the function for surface area of a cone, and I can't seem to extract any more info from the description of the problem.
The first thing I would do is draw a picture! Draw a circle (representing a side view of the sphere) and draw an isosceles triangle in the circle (representing the cone inside the sphere). Use that to find a relation involving h, the height of the cone, r, the radius of the base of the cone, and R, the radius of the sphere.
Oh, okay, I had done it before, but only now realized that I could use the equation of a circle as a constraining factor. And... I got the correct answer.
Thanks. Sometimes it's good to be able to rely on the authority of an experienced person\mathematician while you're still learning the subject.