# Find the utmost value of the volume of cylinders inscribed into a sphere which has R radius.

kjcdb8er | Certified Educator

Answer this question by slicing the sphere into small rings which get larger in diameter as you go from the top of the sphere to the hemisphere. A cylinder is then formed from each of these slices by extending the slice through the sphere until it reaches the bottom boundary.

V = PIr^2h

r = Rcos(t)

(1/2)h = Rsin(t)

--> V = 2PI*R^3 cos(t)^2 sin(t)

Then, differentiate and set V' to zero to find the maximum:

V' = 2PI R^3 [ -2cos(t) sin(t)^2 + cos(t)^3 ] = 0

--> -2sin(t)^2 + cos(t)^2 = 0

--> tan(t) = 1/sqrt(2) or t ~= 35 degrees

The cylinder corresponding to t = 35 has:

r = Rcos(35) and h = 2Rsin(35)

for V = 2PI R^3 cos(35)^2 sin(35)

giorgiana1976 | Student

If we are setting the cylinder's height to the value 2x, the radius of cylinder's base is sqrt (r^2-x^2) and volume of cylinder will be calculated in this way:

V(x)=2*pi*x*(r^2-x^2)

For finding the maxim value of the volume we have to find out the first derivative of the function V(x) and after that, to calculate the roots of the equation V'(x)=0

V'(x)=(2*pi*r^2*x-2*pi*x^3)'=2*pi*r^2-6*pi*x^2

V'(x)=2*pi*(r^2-3*x^2)=0

r^2-3*x^2=0

3*x^2=r^2

x^2=(r^2)/3

x=r*sqrt(3)/3

Vmax is calculated for the maxim value for x=r*sqrt(3)/3

Vmax=2*pi*r*sqrt(3)/3*(r^2-r^2/3)=4*pi*r^3 *sqrt (3)/9