Question

A right circular cone is inscribed in a sphere of radius R. What is the largest possible surface area of the cone ?

Answer 1

draw the circle in 2 dimensions with a right angle triangle inscribed.
there will be a vertical line connecting from the top of the circle to some point past the centre. What we're really trying to figure out is the overall 'height' of this right angle triangle. we'll call this height 'h' and the length of the base of the triangle 'b'

so there will be two right angle triangles here: the big one with b^2 + h^2 =s^2
(we'll call the hypoteneuse of this triangle 's')
the second triangle will be if you draw a line connecting the centre of the circle to the right vertex of the triangle if you have the triangle with the hypoteneuse running from the top to bottom right. h-r is the term describing the length from the centre to the bottom of the triangle.

for terminology, r is the radius of the circle
so now we have:
(h-r)^2 + b^2 = r^2
with algebra, this becomes h^2 - 2hr + b^2 = r^2
and then b^2 = 2hr - h^2

SA for a cone = pi rs + pi r^2
r here is our base, 'b' and 's' is our 's'
so b=square root of 2hr - h^2
s=b^2 + h^2 (from our right angle triangle)

simplify the terms. take dSA/dh
find when it equals zero and that's when you'll have your max/min. there's your relationship between r and h. once you have that, you've related your dimensions of your cone to your sphere and can just plug in the values to find your maximum surface area.