hi everyone today we're going to talk about how to find the largest possible volume of
a cones shaped cup to complete this problem we'll draw a picture of the problem and write
what we know identify optimization and constraint equations and then use the derivative of the
optimization equation to find the volume let's take a look in this particular problem we've
been told that a cone shaped drinking cup is been from a circular piece of paper of
radius r where a sector of the piece have been removed and the edges here diagramed
in this figure c a and c b have been joined together and we've been asked to find the
maximum volume of this cone shaped cup so as with any optimization problem the first
thing we want to do is draw a picture of the problem and write down everything that we
know so in case this picture here has already been given to us of the circular piece of
paper but what we need to imagine is the cone shaped cup that's created from it so we know
that the sector that's been cut out here has this center c theses points a and b the angle
a and we know that the radius of the circular piece of paper is r and we'll denote that
with capital r what we need to realize is that if we join c a and c b if we pull those
two sides together we'll get a cone shaped cup like this that has a side here of r a
height down the center of the cone of h and a base radius of lower case of r since we're
being asked for the maximum volume of this cone shaped cup we know that we're going to
need the equation of the volume of the cone so we've gone ahead and written volume equals
one third pie r squared h which is the equation for the volume of a cone now our next step
is to realize that at the end of this problem we're going to need to get this volume equation
completely in terms of capital r right now we have it in terms of lower case r and h
but because the problem asks us to maximize the volume of this cup that has a radius right
of capital r we know we're going to have to get the volume equation in terms of capital
r only so we're going to have to do a couple of things in order to get to that point the
first of which is that we're going to have to find some way to relate capital r with
lower case r and h the way that we're going to do that is by imagining that we cut a cross
section of this cone if we took like a knife and we cut this cone straight down right we
started from the top and sawed straight through the middle of it we would get a triangle that
looks like this that has the hypotenuse here on the one side of r this height down the
middle here of h and the base just of this half here as lower case r the base radius
of the cone we know from the pythagorean theorem that we can use this triangle to relate these
three values so we have the two sides here r and lower case h lower case r and h and
the hypotenuse of capital r when we have two sides and hypotenuse like that of a right
triangle we know that lower case r squared plus h squared is equal to the hypotenuse
capital r squared so there's our equation that relates the three variables so now we
need to get to work solving this equation here what we'll call the constraint equation
for one of these other variables the goal here is going to be to solve for one of the
other variables and then plug the value back into the volume equation so what we should
do is realize that we have r squared here and our volume our optimization equation and
we have r squared here in our constraint equation since those two are equal it would be awfully
convenient to be able to solve for lower case r squared so what we'll do is we'll subtract
h squared from both sides and we'll get lower case r squared is equal to capital r squared
minus h squared now we can take the value we have on the right hand side r squared minus
h squared and plug that in for lower case r squared in our volume equation so in our
volume equation we'll get one third pi times capital r squared minus h squared times h
now we really have our volume equation in terms of one variable it looks it's in terms
of two variables but it's actually in terms of one variable because capital r is constant
because we've been told that the radius is equal to r r will never change and they could
have just as easily asked us to find the volume of the cup made out of this piece of paper
with radius four but instead they asked us for r it's representing the same thing r here
is a constant so the only variable in this equation that's left is h so what we're going
to do is simplify the volume equation then take it's derivative with respect to h treating
capital r as a constant set it equal to zero and solve for h so let's go ahead and simplify
first what we'll get here is one third pi r squared h when we multiply this coefficient
one third pi times r squared times h then we'll get minus one third pi h cubed when
we multiply the coefficient here by negative h squared by h so now that we have this for
our volume equation we want to go ahead and take the derivative of the volume equation
so we'll get v prime or the derivative of the volume equation remember we're treating
r as a constant so this first term here one third pi r squared h is the same thing right
as basically saying three h one third is a constant pi is a constant and r is a constant
so if we were taking the derivative of three h what we would get is three we basically
take the constant coefficient and front of the h and that would be our derivative in
this case we can take the constant coefficient one third pi r squared and that's our derivative
so one third pi r squared when we take the derivative of negative one third pi h cubed
we'll use the power rule and we'll multiply this three out in front and then subtract
one from the exponent so three times one third will just give us one that will cancel so
we'll be left with just pi out in front and then we'll subtract one from the exponent
to get h squared and now that we have our derivative we can go ahead and set this equal
to zero and solve for h so we will add pi h squared to both sides we'll get one third
pi r squared is equal to pi h squared we'll go ahead and divide both sides by pi and we'll
get pi to cancel that's just a constant and you can see that what we're left with is just
h squared is equal to one third r squared to solve for h we'll go ahead and take the
square root of both sides and we'll see that h is equal to one over the square root of
three r remember that normally when you take the square root you have pay attention to
both the positive and negative value that could result over here on the right hand side
because we're dealing with a three dimensional figure here in actual space if we took the
negative value negative one over the square root of three r this whole value over here
on the right hand side would be negative and that can't be true because if h were negative
this cone wouldn't exist at all so we only have to pay attention to the positive value
now that we've sold for h remember that our goal here is to get back to the volume equation
we want to make substitutions to replace r squared lower case r squared with something
in terms of capital r squared we also want to replace h with something in terms of capital
r so we need to find an equation that relates lower case r squared to r and another equation
that relates h to capital r we already have an equation solved for h in terms of r capital
r it's right here so we'll go ahead and plug one over root three r in for h but we need
to find a value for lower case r squared in terms of capital r so the way that we'll do
that is by using this equation here for lower case r squared so we'll take lower case r
squared is equal to capital r squared we've solved for h squared it's right here so we'll
go ahead and plug in the right hand side here in for h squared so we'll get minus one third
capital r squared so if you can imagine now to find the common denominator if we multiply
this first term by three over three to get that common denominator now we have three
r squared minus one r squared all over three which will just give us r squared equals two
thirds capital r squared now we have an equation for h in terms of capital r and we have an
equation for r squared in terms of capital r so we'll plug both of these into our volume
equation and then simplify to find our final answer now plugging into this volume equation
we'll get volume equals one third times pi for lower case r squared here we'll plug in
two thirds times capital r squared so we'll get two thirds capital r squared and then
we've also solved for h so we'll plug in one over root three times r in for h one over
root three times capital r and now we'll simplify this as much as we can you can see here that
in the numerator we'll get two that'll take care of the one the two and the two we'll
get pi because we have that in our numerator here and then we'll get r to the third which
will take care of this r squared and this r here so we'll get r cubed and then in our
denominator you can see we have three times three gives us nine times root three so essentially
we have nine times the square root of three and that's it that is our final answer for
the maximum volume of this cone shaped cup this way now that we have a volume equation
that's in terms of capital r in the future we could take any value for capital r any
value for the radius of this circular piece of paper plug it in for capital and we would
immediately know the maximum volume since we now have a formula that relates the two
so I hope you found that video helpful if you did like this down below and subscribe
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