Mathematics, 21.01.2020 05:31 cierra3678

Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere of radius 12.a. the radius is .b. the height is

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Answer from: macyfrakes

$r = 4\sqrt{6}\\h=8\sqrt{3}$

Step-by-step explanation:

To find the largest volume of the cylinder in a sphere. We first need to know the volumes of both sphere and cylinder:

$V_{cyl} = \pi R^2 h$

$V_{sph} = \dfrac{4}{3}\pi r^3$

the radius of cylinder is denoted as 'r'

the radius of the sphere is denoted as 'R'

The radius(R) of the sphere is already provided:

$V_{sph} = \dfrac{4}{3}\pi (12)^3 = 2304\pi$

As the cylinder is inscribed in the sphere, we can think of as the cylinder is growing inside the sphere until it touches the surface of the sphere. Notice what is changing? (the volume of the cylinder is increasing, but there's a limit).

So we can say, that the difference between the volume of the sphere and the volume of the cylinder is:

$V = V_{sph}-V_{cyl}$

$V = \dfrac{4}{3}\pi R^3 - \pi r^2 h$

$V = 2304\pi- \pi r^2 h$

We can also observe that as the cylinder inside the sphere grows, the difference(V) between the two volumes decreases. This value cannot be zero since the cylinder cannot exceed the dimensions of the sphere, but it does have a minimum value!

So all we need to do now is just differentiate the above term, BUT there are two variables here, 'r' and 'h'. We first need to find a relation between the two so that we can replace one of them.

so recall that the cylinder is inside the sphere, and it is growing. When it reaches it maximum volume, the corners of the cylinder will surely be touching the surface of the sphere. But more importantly, the cylinder will be fixated at the very center of the sphere.

so as we know already, from the center the radius of the cylinder will be 'r'

but from the center the height will be, h/2

and the distance from the center of the cylinder to the very edge of that touches the surface of the sphere will be, R.

hence, we've found a relation. that these lengths form a right angled triangle. (image given below)

$R^2 = \left(\dfrac{h}{2}\right)^2+r^2$

$12^2 = \left(\dfrac{h}{2}\right)^2+r^2$

$12^2-r^2= \dfrac{h^2}{4}$

$h=\sqrt{576-4r^2}$

We can substitute this into V

$V = 2304\pi- \pi r^2 h$

$V = 2304\pi- \pi r^2 (\sqrt{576-4r^2})$

$V=2304\pi-2\pi r^2\sqrt{144-r^2}$

$\dfrac{dV}{dr}=0-\dfrac{2 \pi r^{3}}{\sqrt{- r^{2} + 144}} - 4 \pi r \sqrt{- r^{2} + 144}$

We need to find where the difference between the volumes is "minimum", hence dV/dr = 0

$0=-\dfrac{2 \pi r^{3}}{\sqrt{- r^{2} + 144}} - 4 \pi r \sqrt{- r^{2} + 144}\\0=\dfrac{6 \pi r \left(r^{2} - 96\right)}{\sqrt{- \left(r - 12\right) \left(r + 12\right)}}\\\\0 = r(r^2-96)\\\\r = 0,r^2=96\\r=0, r= 4\sqrt6\approx9.798$

we can clearly observe that r = 0 is not the right answer, its denoting that maximum value of the difference.

hence, r = 9.798 is the right answer for the minimum value of the difference.

we can find the height using our relation between 'r' and 'h'

$h=\sqrt{576-4(4\sqrt6)^2}$

$h=8\sqrt{3}\approx13.856$

Find the dimensions of a right circular cylinder of maximum volume that can be inscribed in a sphere

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