I cannot seem to find the correlation between having an interval of a radius of a sphere with finding the greatest lateral surface area of a right circular cylinder inscribed in it.
The question goes like this:
Calculate the dimensions of the right circular cylinder of the greatest lateral surface area that can be inscribed in a sphere of radius $6$ inches.
Since the surface area of a cylinder is $2\pi rh$ and since the distance of the sphere's center and the tip of the cylinder is equal to the radius ($6$ inches), then by Pythagorean theorem,
$$\left(\frac h 2\right)^2 + r^2 =36$$
$\frac h 2$ since the center of the sphere is also the midpoint of the cylinder's height.
Then I found $h$:
$$h= 2\sqrt{36-r^2}$$
By this, we can have the equation $A(r)=SA$ of cylinder and
$$A(r)=2\pi r \cdot 2\sqrt{36-r^2}$$
But then my question is what is the connection of this function with the interval $0\le r\le6$?
Thanks
Best Answer
You don't say so explicitly, but your formulas imply that the symbol $r$ represents the radius of the cylinder and $h$ is the height of the cylinder.
There are many differently-shaped cylinders that can fit inside a given sphere. If the radius of the sphere is $6$, near one extreme, you have a very skinny cylinder whose height is very close to $12$ and whose radius is close to zero. Near the other extreme, you have a cylinder that is almost a flat disk with a very small "height" (or thickness) $h$ and a radius that is barely less than $6$.
We can even take the cylinder all the way to the extreme where it is only a degenerate cylinder: $h=12$, $r=0$ describes a line segment that just exactly fits in the sphere, and $h=0$, $r=6$ describes a flat disk that also exactly fits in the sphere.
There's not much point in considering $r < 0$, because what is a cylinder with a negative radius? (Actually there is a way to interpret such a thing, but if you allow, for example, a cylinder of radius $-2$, it's identical to a cylinder of radius $2$; so you have found no new cylinders but you have to rewrite formulas such as $A(r) = 2\pi rh$ because the lateral area of a cylinder is not less than zero. It's easier, and we don't miss any answers, if we consider only cylinders such that $r \geq 0$.)
There is certainly no point in considering a cylinder with radius $r > 6$ if the radius of your sphere is $6$, because there is no way such a cylinder could possibly fit inside the sphere.
You could also come to these conclusions by examining the formula you found, $$A(r)=2\pi r \cdot 2\sqrt{36-r^2},$$ because $r < 0$ gives a negative area (which can't be correct) and $r > 6$ asks you to take the square root of a negative number, which doesn't work. But I think it's a good idea to have reasons based in the original problem statement that say why a variable $r$ should have only a certain possible set of values, because there's no guarantee that every restriction that the problem statement requires will also be enforced by whatever formulas we derive.
So we are OK with $r=0$ (if we accept a line segment as a degenerate cylinder of radius zero), we are OK with $r=6$ (if we accept a circle as a degenerate cylinder of height zero), and we are OK with anything in between--just set $h= 2\sqrt{36-r^2}$, as you discovered, and the cylinder will fit perfectly. But $r < 0$ is useless to consider and $r > 6$ is impossible. So $0 \leq r \leq 6$ describes every cylinder that we could ever want to inscribe in a sphere of radius $6$.