Fun calculus problem!?

Question

I have solved this a couple years ago, but wanted to revisit the solution...

This dude named Wombat has a ball of gold (a perfect sphere actually). He wants to share some of the gold with his buddy, but wants to make the division needlessly complicated ;)
He therefore cuts an inscribed cylinder out of the sphere and is going to give his buddy either the cylinder or whats left over.
If he wants to be generous and give the buddy MORE gold, which should Wombat give buddy: the cylinder or what's left over?

Answer 1

Assuming the cylinder has been cut to maximize the amount of gold in the cylinder, he should give his buddy the cylinder.

Reason:

Let R be the radius of the sphere and let r be the radius of the cylinder. Drawing a right triangle from the center of the cylinder to the side and then from the side to the top edge of the cylinder, we see that R² = r² + (h/2)², so h = 2√(R²-r²). Therefore, the volume of the cylinder will be 2πr²√(R²-r²). We wish to find the maximum volume of the cylinder. Therefore, we take the derivative of the cylinder's volume with respect to r to obtain:

4πr√(R²-r²) - 2πr³/√(R²-r²)

Setting this equal to zero and solving:

4πr√(R²-r²) - 2πr³/√(R²-r²) = 0
4πr(R²-r²) - 2πr³ = 0
4πR²r - 6πr³ = 0
r=0 ∨ 4πR² - 6πr² = 0
r=0 ∨ 4πR² = 6πr²
r=0 ∨ r² = 2/3 R²
r=0 ∨ r = ±√(2/3) R

Of course, r cannot be negative, so r=-√(2/3)R is out, and r=0 is clearly a minimum. Therefore, it follows that the maximum value occurs at r=√(2/3) R. And the volume of the cylinder at this point is:

2π(2/3)R²√(R²-2/3 R²)
4π/3 R² √(1/3 R²)
1/√3 4π/3 R³
1/√3 V (where V is the volume of the sphere)

This is more than half the volume of the sphere, so the cylinder may contain more gold than the remainder. Thus, he should give his buddy the cylinder.

Answer 2

that would of cource depend on the height of the cylinder,radius of the base of cylinder and radis of sphere... and 1 more thing how can one give somebody a cylinder which is inscribed in a sphere???
ya i got the answer
let R = radius on sphere and r of cylinr and h height puting ...i cant write here but by differentiating after solving H2 =R2+r2 and then putting it in vol of cylinder ......solving.....solving
we get maxima when R =underroot3 h then put this in vol of cylinder and u get the answer

Answer 3

The volume of the cylinder inscribed in a sphere whose diameter d is (pi d^3)/8sqrt2
The volume of the sphere is (pi d^3)/6
so the leftover volume is [(pi d^3)/6]-[(pi d^3)/8sqrt2]
the leftover volume=((4sqrt2 - 3)/24sqrt2)pi d^3=~ 0.078 pi d^3
The volume of the cylinder= (pi d^3)/8sqrt2=~0.083 pi d^3
cylinder has larger volume so Wombat should give the cylinder

Answer 4

I would say the part that is left over.

It would have more surface area than the cylinder and thus appearing to be MORE gold.

Answer 5

the Cylinder.