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Question 192419: Determine the dimensions of the cylinder of maximum volume that can be inscribed in a sphere with a radius of 8cm.
The 8cm is the sphere's radius.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
Consider a diameter of the sphere perpendicular to the inscribed cylinder's axis. From the center of the sphere, construct a radius that intersects the circle of intersection between the sphere and one base of the cylinder. We need to derive a function for the volume of the cylinder in terms of the angle between the above described diameter and the constructed radius. Let the measure of that angle be x.
Working this problem will be computationally simpler if we consider the sphere to have radius r for the time being. Given that, the radius of the base of the cylinder is:
 = r\cos{x})
And the area of the base of the cylinder is then:
 = \pi r^2\cos^2{x})
Half the height of the cylinder is:
And then the height of the cylinder is:
 = 2r\sin{x})
Now we can describe the volume of the cylinder:
 = \pi \left(r^2\cos^2{x}\right)\left(2r\sin{x}\right))
Since  = 0) and  = 0) and these are the limiting values for x, if we can find an extreme point for the function on the interval  , we can be assured that this point is a maximum.
Take the derivative of  :
First factor out the constants:
 = 2\pi r^3 \left(\cos^2{x}\right)\left(\sin{x}\right))
Now define  = \cos^2{x}) and  = \sin{x})
Hence
 = 2\pi r^3f(x)g(x))
Now use the Product Rule:
 = 2\pi r^3 f'(x)g(x) + f(x)g'(x))
Use the Chain Rule to derive )
Let 
Then =y = u^2) and =\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=(2\cos{x})(-\sin{x})=-2\cos{x}\sin{x} )
Also:
 = \cos{x})
So:
 = 2\pi r^3 \left(\left(-2\cos{x}\sin{x}\right)\left( \sin{x} \right) + \left(\cos^2{x}\right)\left(\cos{x}\right)\right))
 = 2\pi r^3 \left(-2\cos{x}\sin^2{x} + \cos^3{x}\right))
 = 2\pi r^3\cos{x} \left(-2\sin^2{x} + \cos^2{x}\right))
A local extreme (and here, a local maximum) is at the point where the first derivative is equal to zero, so:
 = 0)
Applying the zero product rule:
or
But  which is not in the feasible interval, , hence we exclude this value as an extraneous root.
To solve:
Use
So:
and
and, again using  to determine:
Substituting these values you can use the volume function to calculate the volume of the maximum inscribed cylinder for the general sphere of radius r:
 = 2\pi r^3 \left(\frac{2}{3}\right)\left(\frac{sqrt{3}}{3}\right)=\frac{4\pi r^3\sqrt{3}}{9})
Notice that the calculation reduces to simply multiplying the volume of the sphere ) by a factor of  .
Of course, for your specific sphere of radius 8, you need to substitute for r. I'll let you do the arithmetic.
John
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