Math problem about cylinders?

Question

A cylinder is inscribed in a sphere of radius 5. Determine the greatest lateral surface area of such a cylinder

Answer 1

say cylinder's radius is r, height is h

then lateral surface S = 2Pi * r * h

from pythagoras,

h^2 + (2r)^2 = 5^2

so h = sqrt (25 - 4r^2)

S =
2Pi * r * h =
2Pi * r * sqrt (25 - 4r^2)

S' =
2Pi [ sqrt (25 - 4r^2) + 1/2 * r * (25 - 4r^2)^(-1/2) * (-8r) ] =
2Pi [ sqrt (25 - 4r^2) - 4 * r^2 * (25 - 4r^2)^(-1/2) ].

To maximize, S' = 0, so

2Pi [ sqrt (25 - 4r^2) - 4 * r^2 * (25 - 4r^2)^(-1/2) ] = 0
sqrt (25 - 4r^2) - 4 * r^2 * (25 - 4r^2)^(-1/2) = 0
sqrt (25 - 4r^2) = 4 * r^2 * (25 - 4r^2)^(-1/2)

square both sides:

(25 - 4r^2) = (16 * r^4) / (25 - 4r^2)

(25 - 4r^2)^2 = 16 * r^4

625 - 200r^2 + 16r^4 = 16r^4

625 = 200r^2
r^2 = 625/200 = 3.125

r = 1.768
h = sqrt (25 - 4r^2) = 3.536

so, surface area =
2Pi * r * h =
39.28

Answer 2

First, symbology:

h = height of cylinder
r = radius of cylinder

A right triangle can be formed using the diameter of a cylinder and its height. Since the cylinder is inscribed inside a sphere, the diameter of the sphere is equal to the diagonal of that right triangle.

So, the right triangle has legs of length h and 2r and a diagonal of 10.

So h^2 + 4r^2 = 100.

The lateral surface area of a cylinder is A = 2(pi)rh.

We need to rewrite the equation h^2 + 4r^2 = 100 in terms of one of the variables. So h = sqrt (100 - 4r^2) = 2*sqrt(25 - r^2).

Thus, A = 2(pi)r[2*sqrt(25 - r^2)]

Take the derivative of this equation with respect to r to find the maximum point.

A' = 4(pi)sqrt(25 - r^2) + [4(pi)r* (-2r)]/[sqrt(25 - r^2)]
0 = 4(pi)(25 - r^2) + 4(pi)r*(-2r)
0 = 100(pi) - 4(pi)r^2 - 8(pi)r^2
0 = 25(pi) - 3(pi)r^2
25(pi) = 3(pi)r^2
r^2 = 25/3
r = 5sqrt(3)/3

The maximum lateral surface area is approximately 148.1 square units.

Answer 3

In order for the cylinder to be inscribed in the sphere, the diagonal of the cylinder (bottom right to top left) is the diameter of the sphere. From this arises a Pythagorean relationship between the cylinder's diameter and height:

(2r)^2 + h^2 = 10^2
h = sqrt (100 - 4r^2) = 2sqrt (25-4r^2)

The formula for lateral surface area of a cylinder is 2*pi*R*H.

A = 2pi* r * 2sqrt (25-4r^2)
= 4pi*r* sqrt (25 - 4r^2)
Derive using product rule:
A' = 4pi*(25-4r^2)^.5 + 4pi*r* [.5*-8r* (25-4r^2)^-.5]
A' = 4pi*(25-4r^2)^.5 - 16pi*r^2* (25-4r^2)^-.5
0 = 4pi (25-4r^2)^-.5 [(25-4r^2 - 4r^2]
0 = 4pi (25-4r^2)^-.5 (25-8r^2)
critical values are r = 5/2 and r = 1.25 sqrt 2

The maximum occurs when x = 1.25 sqrt 2 yielding an area of 25pi.

Answer 4

x = 1.25 sqrt 2