Dimensions that minimize the surface area of a cylinder (KristaKingMath)

Dimensions that minimize the surface area of a cylinder (KristaKingMath)


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 below and subscribe to notified of future
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100 thoughts on “Dimensions that minimize the surface area of a cylinder (KristaKingMath)

  1. She's pretty good looking I spose, about a 7 out of 10. Maybe 8 out of ten if it weren't for that lazy eye. Keep up the good work!

  2. god.. I've got a HUGE head ache over this question, all because I did not convert 1L to 1000cm^3

    But really I don't think it matters what you use as a unit. Does it?

  3. But even still when I follow your steps to similar problems Im not getting the right answer… probably another editing mistake in the calc book… this really makes me wonder, the teams of people making this calc book must have the combined brain power of a turnip.

  4. i divided both sides by 4. since that gives me 0/4 on the left side, i just have 0. i get 4/4 on the right side, which just leaves me with 1. the r^2 disappears when you take the cube root of both sides. 🙂

  5. this… HELPED MEEEEEEEEEEEEEEEEEEE!
    I knew how to do the problem, but I didn't know why I did the problem. now I know WHY!

  6. I fully understand now, thank you 😀 Feeling ready for my exam. This was the only chapter I struggled in

  7. I did this but it gave me a larger SA than the original. I feel that I accidentally maximized it but I don't know how I did this. 

  8. Honestly Calculus is tough. I'm a junior in high school currently in Calculus AB. I would have never thought it would have been this confusing. I was looking for a tutor and you seemed perfect 🙂 , would you consider it?

  9. Thank you!!!! You just saved me so much! I was in a time crunch, and your video was concise, easy to understand, and visually appealing. You have my neverending gratitude.

  10. Hey Krista, how do you know that by solving for r when the equation is set to 0, that will be the absolute min? or is it just understood that that is the absolute min?

  11. Why didn't you take the second derivative to actually check that you had a minimum?  The assumption was that you had a minimum, and it's true, but you did not show how to check that.  How come?  Thank you.

  12. Thank you for posting this. Your explanation was very clear and I was able to follow along so I could get my project done. 

  13. how could you dislike this video is the only puzzling thing about it… Thanks alot for the post it's always really helpful 😀

  14. very helpful video! now to find the height that would minimize the amount of material for the cylinder or can used would I just plug in the radius I got as the final answer into the height equation??

  15. Very helpful video! But are the measurements for radius and height in cm or inches? Doing a project that requires a literal construction of the container

  16. i believe that DJ khaled once said about this page and I quote: "You Smart. You Real Smart. Matta fact baby you's a genius." #keystosuccess

  17. Muchas Gracias, todo se puede lograr. Amo la Ciencia y la Tecnologia, Siento que todo se puede lograr con tus Clases. Gracias Maestra por su tiempo. ES POSIBLE. Matemáticas un gran misterio.

  18. What if you were asked to minimize the Energy of a particle in it's ground state in a cylinder for a given volume?

  19. question: I am doing one right now thats the same question except the cylinder has no top. Would I simply subtract (pi)r^2 from the optimization equation or is there more to it then that?

  20. hey, i have a doubt, if the derivative gets the maximum value instead of minimum? will we still be able to solve to get minimum value?

  21. I'm struggling to find what my optimization and constraint is. I've been told the cylinder has to have volume V and I need to know the shape of the can that is the most efficient cost and cost is equal to $1/per square cm.
    Would my optimization be the minimal cost of the can? Shape of the can is my constraint?

  22. Hi, I am looking for a formula to calculate the lenght of a rope enrolled flat on a deck. Kind of spiral enrollment every boat lover have seen.

  23. Great explanation!! It only occurred to me recently that there ought to be a formula for minimizing the surface area of a cylinder, both to reduce construction costs, but also to limit radiation heat loss or gain. Moreover, the height to radius should be ratios, scalable to any size. I love geometry and trig, but never a big fan of calculus. Thanks for a fairly intuitive explanation. I was planning on using a spreadsheet for trial by error or range.

  24. I saw another optimization problem which was asking for the maximization.. in that problem once they found the critical point.. they had to test which interval gives a positive value. So my question is how do u know that answer is the one that minimizes? How come u don’t have to test any interval once u get ur critical point (radius) ??

  25. Would also be interesting to show in the conclusion to the video that area is minimized when h=2r (when the height of the cylinder equals its diameter). Just an interesting observation.

  26. Using your solutions ..The ht is twice the radius …h= 10.8 .,,,,.r=  5.4Will the height always be twice the radius when it asks for dimensions that minimize  Surf area of a cylinder?

  27. You smart, but y is it that when i plug my dimension values into the volume equation i ain't getting 1000 cm^3?

  28. a sphere of radius 5 cm is placed on a cylinder of a base of 4 cm find out the volume of part of the square that lie inside the vessel

  29. I was wondering what was the working out between the simplified area and area prime. how was the derivative formed from the equation before?

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