A right circular cylinder is inscribed in a sphere of radius r.?

Question

Find the largest possible surface area of such a cylinder. (Show your work.)

Answer 1

Let the radius of the sphere be R.
Let x be the radius of the cylinder.
Then height of half of the cylinder= sqrt(R^2-x^2)

Total surface area of cylinder = 2*pi*x^2 +2*pi*h, so
S=2*pi*x^2 +2*pi*2*(R^2-x^2)^.5 [Eq I]
dS/dx=4*pi*x + 4*pi*.5*(-2x)(R^2-x^2)^-.5
=4*pi*x - 4*pi*x/(R^2-x^2)^.5
=4*pi*x(R^2-x^2)-4*pi*x =0
R^2-x^2-1=0
x^2=R^2-1
x=+ or - sqrt (R2-1)
Plug these values of x into [EQ I] to get answer.

Answer 2

Let "r" is radius of cylinder and let b is its height. Volume of cylinder is then: V = r^2 b pi We can spot two similar triangles: 1. triangle formed by axis, generatrix of a cone and radius of a cylinder "r"; 2. triangle formed by axis, generatrix of a cone and radius of a cone R. (The perimeter of the base of a cone is called the directrix, and each of the line segments between the directrix and apex is a generatrix of the lateral surface.) Because these triangles are similar, we can use proportion: (h - b)/b = r/R b = h (1 - r/R) put this in equation for volume of cylinder: V = r^2 h pi (1 - r/R) V = h pi (r^2 - r^3/R) Height of a cylinder "b" has been eliminated from equation, and now volume of cylinder is function of variable "r" (h and R can be considered as constants). To find extreme value (maximum) of that function, differentiate and equalize to zero: dV/dr = 0 dV/dr = h pi (2r - 3r^2/R) = 0 We can simplify solving of this quadratic equation since we know that neither "h" nor "r" equals 0. So we can divide by (2rh pi) to get 1 - (3/2) r/R = 0 r = (2/3)R Earlier we found b = h (1 - r/R) so b = h (1 - (2/3)R/R) = (1/3)h