# WHat is the largest cylinder of radius, #r# and height #h# that can fit in sphere of radius, #R#?

The maximum volume of the cylinder is found if we choose

# r = sqrt(2/3)R # , and#h = (2R)/sqrt(3) #

This choice leads to a maximum cylinder volume of:

# V=(4pi R^3)/(3sqrt(3)) #

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Imagine a cross section through the centre of the cylinder, and let the cylinder have height

#V=pir^2h#

The radius of the sphere,

# \ \ \ \ \ R^2=r^2+(1/2h)^2 #

# :. R^2=r^2+1/4h^2 #

# :. \ r^2=R^2-1/4h^2 #

We can substitute this into our volume equation to get:

# \ \ \ \ \ V=pir^2h #

# :. V=pi( R^2-1/4h^2 )h #

# :. V=pi R^2h-1/4pih^3#

We now have the volume,

# (dV)/(dh)=pi R^2-3/4pih^2#

At a minimum or maximum,

# pi R^2-3/4pih^2 = 0 #

# :. \ \ \ \ \ \ \ \ \3/4h^2 = R^2 #

# :. \ \ \ \ \ \ \ \ \ \ \ \h^2 = 4/3 R^2 #

# :. \ \ \ \ \ \ \ \ \ \ \ \ \ \h = sqrt(4/3 R^2) " "# (obviously we want te +ve root)

# :. \ \ \ \ \ \ \ \ \ \ \ \ \ \h = (2R)/sqrt(3) #

With this value of

# \ \ \ \ \ \ r^2 = R^2-1/4 4/3 R^2 #

# :. \ r^2 = R^2-1//3 R^2 #

# :. \ r^2 = 2/3R^2 #

# :. \ \ r = sqrt(2/3)R #

We should check that this value leads to a maximum (rather than a maximum) volume, We do this by looking at the second derivative:

# (dV)/(dh)=pi R^2-3/4pih^2#

# :. (d^2V)/(dh^2) = -6/4pih #

And as

Hence, the maximum volume of the cylinder is found if we choose

# r = sqrt(2/3)R # , and#h = (2R)/sqrt(3) #

With this choice we get the maximum volume as;

# V=pi R^2((2R)/sqrt(3) ) -1/4pi((2R)/sqrt(3))^3 #

# :. V=(2pi R^3)/sqrt(3) - 1/4pi((8R^3)/(3sqrt(3))) #

# :. V=(2pi R^3)/sqrt(3) - (2piR^3)/(3sqrt(3)) #

# :. V=(4pi R^3)/(3sqrt(3)) #

And obviously the volume of the Sphere is given by:

#V_s = 4/3piR^3#

This is a very famous problem, that was studied by Greek mathematicians way before Calculus was discovered. An interesting property is the ratio of the volume of the cylinder to the volume of the sphere:

# V/V_s = ((4pi R^3)/(3sqrt(3))) / (4/3piR^3) = 1/sqrt(3) #

In other words the ratio of the volumes is completely independent of

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The largest cylinder of radius r and height h that can fit in a sphere of radius R has a diameter equal to the sphere's diameter, which is 2R. Therefore, the cylinder's height h should be equal to the sphere's diameter, 2R, and its radius r should be equal to half of the sphere's diameter, R.

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The largest cylinder of radius ( r ) and height ( h ) that can fit inside a sphere of radius ( R ) will have its height equal to the diameter of the sphere, ( 2R ), and its radius equal to ( r ). Therefore, the maximum dimensions of the cylinder are:

Radius ( r ): The radius of the cylinder is equal to the radius of the sphere, ( r = R ).

Height ( h ): The height of the cylinder is equal to the diameter of the sphere, ( h = 2R ).

So, the largest cylinder that can fit inside a sphere of radius ( R ) has a radius equal to ( R ) and a height equal to ( 2R ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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