hi everyone today we're going to talk about how to find the dimensions of the cylinder
that minimize its surface area 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 dimensions let's take a look in this
particular optimization problem we've been told that a cylindrical can must be made to
hold one liter of oil and then we've been asked to find the dimensions of the can that
will minimize the cost of the metal used to make it as with any optimization problem the
first thing you want to do is draw a diagram of what you've been told and write down as
much information as you can so we've drawn a diagram of the cylinder and we've indicated
r as the radius and h as the height of the cylinder we've also written down that one
liter is equal to a thousand cubic centimeters and that the area formula for a cylinder is
two pie r h plus two pie r squared we also know that the volume of a cylinder is pie
r squared h so really quickly in case you don't remember those remember that the area
formula is two pie r which is the circumference of the circle times h the height of the can
so that gives you the outside or the material used to make the side and then pie r squared
is the area of the circle so that will give you the top and bottom you multiple by two
to account for the top and bottom together so when we simplify we get two pie r h plus
two pie r squared the formula for the volume is the area of the circle times the height
of the can so the area of the base times the height which will give us the total volume
of the can once we've written down all the information we've been given our next step
is to identify a constraint equation and an optimization equation so with any optimization
problem like this you're always going to need an equation you can optimize but you'll also
need a constraint equation because we've been asked to minimize the cost of the metal that
means that we're going to be minimizing the surface area of the can because the metal
is defined by the surface area so minimizing the surface area means that our optimization
equation will be the area equation so let's go ahead and say that this area equation here
will be our optimization equation our constraint equation is the equation that limits us so
we've been told that cylindrical can has to be made to hold one liter of oil that's the
only constraint we've been given so what the problem is asking is for us to optimize the
area the surface area minimize the cost of metal or minimize the cost of the surface
area within the constraint or within the condition that the volume has to hold one liter or a
thousand cubic centimeters so this volume equation here will be our constraint equation
our next step is to figure out how to reduce our optimization equation to just one variable
so because we have in our optimization equation here both r for the radius and h we need to
get rid of one of those the way that we're going to get rid of one of them is by solving
the constraint equation for one of the other and then plugging in to the optimization equation
so let's go ahead and solve our constraint equation our value equation for either r squared
or h in this case let's go ahead and solve it for h because that will be really easy
to plug in to the area equation we just got h right here so the way that we'll do that
is just by plugging one thousand cubic centimeters in for volume we know that has to be the volume
and we'll set that equal to pie r squared h and we'll solve for h obviously to solve
for h we'll just divide both sides by pie r squared so we'll get one thousand divided
by pie r squared now that we've solved for h we can go ahead and plug that back into
our area equation so if we say area over here is equal to two pie r multiplied by h which
we solved for as one thousand over pie r squared plus two pie r squared now we have our optimization
equation for surface area in the form of one variable r and now that we've got it in one
variable all we need to do is simplify take the derivative set the derivative equal to
zero and solve for r and r should give us the radius of the can that minimizes the surface
area so let's go ahead and simplify this first we'll see that we get pie to cancel here and
we'll get r to cancel from the numerator and the squared to cancel in the denominator so
we'll be left just two thousand divided by r plus two pie r squared let's go ahead now
we need to take the derivative because we've simplified this as much as possible so let's
go ahead and make this a little easier on ourselves we'll call this two thousand r to
the negative one remember we can move a variable from the denominator to the numerator just
by changing the sign in the exponent from a positive to a negative so two thousand r
to the negative one plus two pie r squared if you're not comfortable doing this you can
obviously use quotient rule to figure this out so we're going to take the derivative
a prime and we'll get negative two thousand r to the negative two plus four pie r and
when we move this r to the negative two back to the denominator we'll get negative two
thousand and actually let's go ahead and make that our second term so we don't lead with
the negative we'll get four pie r minus two thousand over r squared so that is our derivative
equation now we need to set this equal to zero and solve for r so if we set it equal
to zero let's at the same time go ahead and find a common denominator in order to find
a common denominator we're going to be multiplying this first term by r squared over r squared
and we'll get four pie r cubed we can combine the fractions minus two thousand all over
r squared so now what we can see is that this right hand side here is going to be equal
to zero whenever the numerator is equal to zero or if the denominator is undefined however
if the denominator is undefined that means r is equal to zero and we know r can't be
equal to zero because we're talking about a three dimensional here in real space and
if r is equal to zero then the cylinder doesn't exist so we can almost just discount that
and we're really just looking at where the numerator is equal to zero so let's go ahead
and factor out a four and we'll get four times pie r cubed minus five hundred obviously now
we can see that the right hand side will equal zero only when pie r cubed minus five hundred
is equal to zero so we can really just ignore the four or divide both sides by four and
that will go away and now if we add five hundred to both sides we'll get five hundred equals
pie r cubed if we divide both sides by pie we'll get r cubed equals five hundred divided
by pie and if we solve for r by taking the cube root of both sides we'll get r equals
the cube root of five hundred divided by pie we know now that this must be the radius that
minimizes the surface area of the can if we want to find the height as well we can just
go ahead and plug r into our height equation it's not very clean but we'll get one thousand
divided by pie times and now we have r squared here so we get the cube root of five hundred
divided by pie all squared that's really just going to give us one thousand divided by pie
times five hundred over pie to the two thirds power so these are the dimensions h here and
r here that minimize the surface area and therefore minimize the cost of the metal used
to construct the can so I hope you found that video helpful if you did like this video down
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