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In geometry, the inscribed sphere or insphere of a convex polyhedron is a sphere that is contained within the polyhedron and tangent to each of the polyhedron's faces. It is the largest sphere that is contained wholly within the polyhedron, and is dual to the dual polyhedron's circumsphere.
The radius of the sphere inscribed in a polyhedron P is called the inradius of P.
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Largest possible volume of a cylinder inscribed in a sphere (KristaKingMath)
Max volume of a rectangular box inscribed in a sphere (KristaKingMath)
Cone Inscribed in Sphere - Part I.wmv
Transcription
hi everyone today we're going to talk about
how to find the volume of the largest cylinder
that will fit inside a sphere to complete
this problem we'll draw a picture of the problem
and write what we know identify optimization
and constraint equations and the use the derivative
of the optimization equation to find the volume
let's take a look in this particular problem
we've been asked to find the largest possible
volume of a right circular cylinder that can
be inscribed in a sphere with radius r so
as with any optimization problem the first
thing we want to do is draw a picture of what
we know so we've been told we're inscribing
a cylinder into a sphere that means that we're
trying to fit the cylinder inside of sphere
so I've draw the sphere here in blue and drawn
the cylinder in orange we also want to label
every part of this diagram that we can so
we know that radius of the sphere has to be
r right we've been told that the sphere has
to have radius r so we've gone ahead and labeled
the radius of the sphere as r we also know
that since this is a right circular cylinder
that we're going to raids of the cylinder
so we can go ahead and call that capital r
and we've labeled that as well and finally
the other thing we know at this point is that
the cylinder is going to have a height obviously
we don't know what it is yet but we know that
its going to have a height and we can call
the height h so we've drawn a picture of that
and then we've also gone ahead and written
down the formula for the volume of the cylinder
we know that we're going to have to find the
largest possible volume of the right cylinder
so we can say that the volume of the cylinder
v sub c is equal to pie capital r squared
h capital r squared because we're talking
about the volume of the cylinder and we've
indicated the radius of the cylinder is capital
r so now that we've drawn a picture of the
problem and written down what we know we need
to identify optimization and constraint equations
remember that optimization equation will be
the equation that we're trying to optimize
meaning the equation that we're trying to
minimize or maximize since we were asked to
find the largest possible volume that means
our optimization equation has to be an equation
for volume so this volume equation we have
here for the volume of the cylinder will be
our optimization equation we also though need
a constraint equation and the constraint equation
is the equation that limits us so we're trying
to maximize the volume as much as we can within
the constraint that the radius of the sphere
must be r that's the only condition that we've
been given the only constraint within which
we have to work so we know that the constraint
equation has to be an equation involving r
and keep in mind this is lower case r for
the radius of the sphere as it turns out we
don't have an equation yet that involves lower
case r or the radius of this sphere so we
need to find one more specifically we need
to find an equation for the radius that is
in terms of capital r the radius of the cylinder
and h so that we can relate lower case r capital
r and h so the way that we're going to do
that is by imagining that we take a cross
section of this sphere that we've diagramed
up here so if you can imagine if we cut out
a slice of this sphere vertically and we had
a vertical slice like this we'd have the height
h here we'd have the radius of the sphere
diagramed here with lower case r and we'd
have the radius of the cylinder diagramed
here with capital r the key to relating these
three variables is realizing that the radius
of the sphere is r everywhere so not only
do we have the radius of the sphere equal
to r here but obviously from the center of
the circle to this point here the radius is
also r so the reason that this is so great
is because once we realized that this length
from the center of the sphere and the cylinder
to the corner of the cylinder or the edge
of the sphere is r we realize that we have
a right triangle and we know that we can relate
the three sides of the right triangle with
the pythagorean theorem remember that the
pythagorean theorem is just a squared plus
b squared equals c squared where a squared
and b squared are just two sides of the triangle
c squared being the hypotenuse of the triangle
we already know that the hypotenuse of the
triangle is lowercase r or the radius of the
sphere so we can say that this is a squared
plus b squared equals r squared and we just
need to plug in the other two sides we know
that one of the sides is capital r we know
that the other side is actually h over two
because the height of this rectangle here
is h but the height of this portion right
here is just h over two so we'll plug in h
over two for a and square it we'll plug in
for b and square that and then we already
know that that's equal to the hypotenuse lowercase
r and that's squared so now we have an equation
that relates lowercase r capital r and h and
this our constraint equation because remember
our constraint equation had to involve lowercase
r now once you have a constraint equation
and an optimization equation you're goal will
be to solve the constraint equation for one
of the variables in your optimization equation
so that you can make a substitution you need
to get your optimization equation in terms
one variable only right now we have two variables
involved in our optimization equation capital
r and h so we need to eliminate one of them
and the way that we're going to do that is
by solving the constraint equation for one
and making substitution so let's go ahead
and solve for r squared as you can see we
have r squared right here and we also have
r squared up here so if we solve for r squared
in our constraint equation we can make a really
easy substitution as opposed to having to
solve for h which may be a little bit more
difficult so let's go ahead and solve for
capital r squared the way that we'll do that
is by taking the square of h over two and
we'll get h squared over four plus capital
r squared is equal to lower case r squared
when we subtract h squared over four from
both sides we'll get capital r squared equals
lower case r squared minus h squared over
four so now we've solved for r squared and
we can go ahead and plug the whole right side
here in for r squared up in our volume equation
when we do that we'll get the volume for the
cylinder is equal to pi times here's where
we're plugging in r squared minus h squared
over four all multiplied by h now our volume
equation or optimization equation is in terms
of one variable only you might think it's
in terms of two variables because you see
lower case r and h however remember that lower
case r is actually a constant because it's
the radius of the sphere which is never changing
and we've been asked to find the largest possible
volume of the cylinder inscribed in the sphere
with radius r they could've just as easily
asked us to find the largest possible volume
of the cylinder inscribed in the sphere with
radius four if they had done that r obviously
would be a constant it would be four in this
case it represents exactly the same thing
we're going to treat it as a constant and
so that means that our volume equation here
only has one variable and that's h r will
be a constant so now that we have our volume
equation or our optimization equation in terms
of one variable only our next step will be
to simply as much as possible and then find
the derivative of the volume equation so let's
go ahead and simplify first we're going to
multiply each term inside here r squared minus
h squared over four by both pi and h simplifies
so we'll get the volume of the cylinder is
equal to we have pi times r squared times
h so we'll get pi r squared h and then we
have pi times negative h squared over four
times h so we'll get minus pi over four h
cubed so that's as simple as we can get our
volume equation now we want to go ahead and
take the derivative of the volume equation
and then set it equal to zero so the derivative
we can call the derivative of v with respect
to h or dv over dh we're going to take the
derivative of the right hand side with respect
to h keeping in mind that r is a constant
so if we take the derivative of the first
term pi r squared h remembering that r squared
is a constant the derivative is just pi r
squared because this is the same here if h
is our variable as taking the coefficient
here pi r squared it's the same thing as taking
the derivative of three x for example where
x is the variable and three is the constant
coefficient pi r squared h sense r is a constant
that whole thing is the constant coefficient
so the derivative of pi r squared h is just
pi r squared same thing here when we take
the derivative with respect to h we'll multiply
we'll use the power rule and we'll bring the
three out in front and subtract one from the
exponent so we'll get minus three pi over
four h squared so now that we've got our derivative
we want to go ahead and set it equal to zero
and solve for h so in order to solve for h
let's go ahead and add three pi over four
h squared to both sides we'll get pi r squared
is equal to three pi over four h squared let's
go ahead and multiply by both sides by four
we'll get four pi r squared is equal to three
pi h squared and now we can go ahead and divide
both sides by three pi when we do that pi's
will cancel and we'll just be left with four
r squared divided by three is equal to h squared
to solve for h we'll take the square root
of both sides and when we take the square
root of both sides we can take the square
root of the numerator and the denominator
of the left hand side separately so the square
root of four r squared is just two r and the
square root of three we'll go ahead and leave
as the square root of three and that will
be equal to h now that we have a value of
h we just need to return to our original problem
and remember what we were actually asked to
find remember that we asked to find the largest
possible volume of the cylinder so we need
to go ahead and plug in our value for h into
our volume equation this was the last good
volume equation that we had so let's go ahead
and plug two r over the square root of three
into this volume equation for h when we do
that we'll get the volume of the cylinder
is equal to pi r squared times two r over
the square root of three minus pi over four
times two r over the square root of three
cubed now that we've plugged in h our next
step is just to simplify the volume equation
as much as possible so we get the volume of
the cylinder is equal to two pi r cubed divided
by the square root of three minus pi over
four when we cube two r over the square root
of three we get eight r cubed divided by three
root three when we distribute the pi over
four across the eight r cubed three root over
three root three we'll get we'll get the four
in the denominator to cancel and the eight
will become a two and so what we'll get is
two pi r cubed divided by three root three
now in order to find the common denominator
all we need to do is multiply this first term
here by three over three so we get six pi
r cubed minus two pi r cubed all divided by
three root three obviously that just gives
us four pi r cubed over three root three and
that's it that's actually the largest possible
volume of the right circular cylinder that
can be inscribed in this sphere radius r now
that we have this equation what it would allow
us to do is immediately find the largest possible
volume of the cylinder regardless of the radius
r that we're given so for example if we were
told that the radius of the sphere would be
equal to two we could plug in two for r we'd
get eight up here on top we'll get thirty
two pi and we'd know immediately that a largest
possible volume of the cylinder or the volume
of the cylinder if the radius is two is thirty
two pi divided by three root three so no matter
what we get for r we can find the largest
possible volume of the cylinder given this
formula but this is our final answer for the
largest possible volume when the radius of
the sphere is r so I hope you found that video
helpful if you did like this video down below
and subscribe to be notified of future videos
All regular polyhedra have inscribed spheres, but most irregular polyhedra do not have all facets tangent to a common sphere, although it is still possible to define the largest contained sphere for such shapes. For such cases, the notion of an insphere does not seem to have been properly defined and various interpretations of an insphere are to be found:
The sphere tangent to all faces (if one exists).
The sphere tangent to all face planes (if one exists).
The sphere tangent to a given set of faces (if one exists).
The largest sphere that can fit inside the polyhedron.
Often these spheres coincide, leading to confusion as to exactly what properties define the insphere for polyhedra where they do not coincide.
For example, the regular small stellated dodecahedron has a sphere tangent to all faces, while a larger sphere can still be fitted inside the polyhedron. Which is the insphere? Important authorities such as Coxeter or Cundy & Rollett are clear enough that the face-tangent sphere is the insphere. Again, such authorities agree that the Archimedean polyhedra (having regular faces and equivalent vertices) have no inspheres while the Archimedean dual or Catalan polyhedra do have inspheres. But many authors fail to respect such distinctions and assume other definitions for the 'inspheres' of their polyhedra.
This page was last edited on 19 February 2021, at 05:35