Calculus optimization help! A right circular cylinder is inscribed in a sphere of radius r. Find the dimension?

I'm going to start by just using 2-dimensional space, a circle (so we don't have to worry about getting rid of the "z" term).

A circle, centered at the origin with radius r, has an equation of: x^2+y^2=r^2

If you superimpose the cylindar in question on this circle, the radius from the center of our circle, to the top-right corner of the cylindar, will have coordinates of (x,y) at that top-right corner.

The height of our cylindar is just 2*y

The radius of the cylindar is just x

So the surface area (assuming both top & bottom surfaces are included) is:

S.A. = 2*(top surface area) + side surface

S.A. = 2πx^2 + (2πx)*(2y)

S.A. = 2πx^2 + 4πxy

We can only differentiate with a single variable, so we need to get rid of the "y" in that last equation. We have our original equation, x^2+y^2=r^2. So y=sqrt(r^2-x^2)

S.A. = 2πx^2 + 4πxy

S.A. = 2πx^2 + 4πxsqrt(r^2-x^2)

Now take the derivative:

S.A.' = 4πx + 4π(x*(1/2)sqrt(r^2-x^2)^(-1/2)*(-2x) + sqrt(r^2-x^2)) = 0

Multiply everything through by sqrt(r^2-x^2), and divide everything by 4π:

x + (x*(1/2)sqrt(r^2-x^2)^(-1/2)*(-2x) + sqrt(r^2-x^2)) = 0

xsqrt(r^2-x^2) - x^2 + r^2-x^2 = 0

xsqrt(r^2-x^2) =2x^2 - r^2

square both sides

x^2(r^2-x^2) = 4x^4 - 4x^2r^2 + r^4

x^2r^2-x^4) = 4x^4 - 4x^2r^2 + r^4

5x^4 -5x^2r^2 + r^4 = 0

Using the quadratic formula, with:

a=5

b=-5r^2

c=r^4

x^2 = (5r^2 +/- sqrt(25r^4 - 20r^4))/10

x^2 = (5r^2 +/- sqrt(20r^4))/10

x^2 = (5r^2 +/- 2r^2sqrt(5))/10

x^2 = r^2(5 +/- 2sqrt(5))/10

x = r*sqrt((5 +/- 2sqrt(5))/10)

this becomes (using excel):

x = 0.973249r, x = 0.229753r

so y=sqrt(r^2-0.947214) or y = sqrt(r^2 - 0.052786)

Hello Carl, let R be the radius of the sphere and r be the radius of the cylinder with height 2h. Just for mathematical convenience Then lateral surface area is 4 pi r h As you draw the diagram, then you can get R^2 = r^2 + h^2 A = 4 pi ./(R^2 - h^2) * h dA/dh = 0 as A has to be maximum Hence differentiating, dA/dh = 4pi *[h * 1/2./(R^2 - h^2) * (-2h) + ./ (R^2 - h^2)] Equating this to 0, we get, R^2 - 2h^2 = 0 Or h = R /./2 Hence the height of cylinder with max lateral surface area is (./2) R So radius will be r = R /./2 So max area = 2 pi R/./2 * ./2 R = 2 pi R^2 Max lateral area = 128 pi sq units.