The whole problem turns on one picture. Slice the sphere straight down through the axis of the cylinder, and the cylinder becomes a rectangle sitting inside a circle of radius 8, with all four corners touching the circle.
Draw the radius from the center of the sphere out to one of those corners. That radius is the hypotenuse of a right triangle whose legs are the radius of the cylinder, r, and half of the cylinder height, h/2, since the center sits halfway up. So by the Pythagorean theorem
r2 + (h/2)2 = 82 The corner of the cylinder lies on the sphere
r2 = 64 − h2/4 Solve for r2
That is the relationship that lets us get rid of r. We never actually need r by itself, only r2, which is lucky because r2 is exactly what shows up in the volume formula:
V = πr2h Volume of a cylinder
V(h) = π(64 − h2/4)h Substitute what we found for r2
V(h) = 64πh − πh3/4 Distribute
So that is the equation they asked for, the volume written only in terms of h.
Now to maximize it. Take the derivative and set it to zero, since the volume is largest where the slope flattens out:
V'(h) = 64π − 3πh2/4 Differentiate term by term
64π = 3πh2/4 Set it equal to zero and move the second term over
256 = 3h2 The π cancels, then multiply both sides by 4
h2 = 256/3
h = 16/√3 ≈ 9.24 inches
We take the positive root because h is a length. Putting that back into the volume gives
V = 64π(9.24) − π(9.24)3/4
V ≈ 1238.22 cubic inches
So the height that maximizes the volume is about 9.24 inches and the maximum volume is about 1238.22 cubic inches. The exact value is 2048√3π/9 if your teacher wants it unrounded.
One thing worth noticing is that the tallest cylinder is not the biggest one. If you push h up toward 16 the cylinder becomes a thin needle and r shrinks to nothing, and if you push h down toward 0 it becomes a flat disc. The best one sits in between, and because the volume depends on r2 but only on h to the first power, the trade favors keeping the cylinder fairly wide.
