Optimization Problem #6 - Find the Dimensions of a Can To Maximize Volume
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- Опубликовано: 16 сен 2024
- Thanks to all of you who support me on Patreon. You da real mvps! $1 per month helps!! :) / patrickjmt !! Optimization Problem #6 - Find the Dimensions of a Can To Maximize Volume. In this video, we have a certain amount of material with which to make a cylindrical can. We have to find the radius and height that would maximize the volume of the can.
you have a great math voice, great handwriting and a gift for teaching math! Thanks for sharing!
Notice that his final result (height = ~4 and radius = ~2) is basically a cylinder with the same height as it has diameter. The optimal dimensions of shapes after being optimized is nearly always as close to a square (or cube) as you can get, and a cylinder having equal height and diameter is basically a cube-like cylinder :)
Just a way to check your sanity when doing these problems.
Our teacher told us that we need to find an optimization word problems for our performance exam in our basic calculus. I found this problem and this is what I gave to him. Today is our exam for basic cal, i just finished presenting this problem. At first, i was really tensed because i think i might get it wrong. Our teacher said we are not allowed to bring our copies in front so what I/we did is we memorized our different problems and it's solutions. Thank you so much for this video! I don't know what will I present if it wasn't for this. THANK YOU SO MUCH.
(I added a comment cause i want u to know that i appreciate this video, Thank you so much!)
you are very welcome :)
Thanks a ton! We are currently doing a project on maximizing volume in a soda can and other cylinders. This helps so much!!
do you happen to know why we set it equal to 0 after finding the derivative ?
@@sarahameed114 to find the maximum values
well, in this case, i suppose that technically that the radius could be arbitrarily large, so long as we made the can arbitrarily short! so, the radius is not really bound by any specific number.
I wish you were my math teacher, these videos are so helpful!
@ArcHeus08 if you want to clearly justify, that would not be a bad thing to do! (can also use the second derivative test)
This was one of the topics I was confused about in my Calculus class. I have a final on Sunday and you sir really helped me out. Thanks man. I rely on your videos a lot now because it's more application than conceptual. I would love to know how to prove to come up with the concept, but due to tests, the application of the problems is priority. Most other youtube videos are proving the concepts.
It looks hard at first, but you CAN do it!
LAME
Thank you so much this was extremely straightforward and helpful
i like your work sir. it really helped me a lot. i need your help on this question. " A box with an open top is made from a rectangular piece of cardboard, by cutting out squares in the corners of the cardboard and folding the sides upwards. The cardboard measures 50 by 30 cm. calculate the length of the side of the squares to be cut out to give a maximum volume"... please i trust you on this.
Wow..you saved my life..!!Thank you so mcuh!
Much more satisfying if you leave S (surface area of given material) unknown and maximise V for a given S, you find that diameter = height
Thanks Sir
Thanks for the video.
At 6:11 when you take the derivative how come you don't have to write the radius as Dr/dt?
@paulbeinar * I MUST COMPUTE *(in robot voice)
When do you know to find the derivative? After you finish simplifying?
you saved my lifffffeeeee :)
glad i could help :)
@@patrickjmt but I have a question tho , in V 1 from where did we get the 3 in front of pi r^2
Thank you for this video!
Amazing video thank you!
If you leave the rounding to the very end, I think you'll find the height should be 4.13inches to 3sf.
So deriving V helps us find the radius specific to the cylinder? Someone confirm
love you so much
Hi Patrick, thanks for this wonderful video. I have a question though; don't we need to use the first derivative test upon getting the value of r to be sure that it is a maximum value? Thanks
It is just bad didactic technique no matter the nastiness of the problem the student will face.
To introduce a concept and for early practice you don't want working memory lost to heavy arithmetic and digit span tasks. It also slows down the presentation making it harder for students to focus.
Once well practiced and with ample time. i agree it is good to use real world nasty numbers.
Because textbooks and tests don't use easy numbers. If you practice with difficult and otherwise confusing numbers, you'll do much better on any examination involving these types of problems.
Why didn't you check that you had a maximum by taking the second derivative?
it is taken in terms of r rather than x.
at 3:49 why don't you just divide evrything by pi to get:
25.625 = 2r^2 + 2rh
?
why not just use A for the area
25.625 pi? Really? Why not make the surface area 1 or some general constant like C. It will save you a lot of writing.
where can i find that type of paper?
+ukeeba1 Printing paper.. cheap one.
+ukeeba1 you will need a master ball to obtain paper of such quality
I love math. Calculus makes physics much easier.
when u took the derivative and solved for r... u never checked to see if it was a local max or min
its impossible for it to be a min, since the only minimum possible volume is zero, which would be at zero height or radius.
but in general yes
i tried working the question first before viewing the answer and i said the perimeter of the cylinder should be 25 625pi=2pi R^2 + h please how did u cum up with 2piR
A circle is a regular polygon with infinitely many sides.
So the perimeter of a circle is the same as the circumference of that circle and is always 2πr
L=2r.
شــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــكرا
@gouzbekistan1 Because he's getting h by it self.
i find this hard to follow. Maybe am just dumb :(
For optimization, you generally need to know the area of something, and the volume of it as well, (or perimeter and area). Whatever you are trying to solve for you must take the derivative of! In this case you wanted to know Volume, so after finding the expression for height (h), Patrick simply plugged height into his volume formula, simplified, then took the derivative of Volume (what you want to optimize) and solved for radius. Now that you have radius, r, you can simply plug that back into your height expression and find r! At which point you now have the optimal Volume for the cylinder! Know your areas, perimeters, and volume equations!
Daniel Lau Go easy on yourself man. Relax a bit and get back on track ... step by step ;) Good luck!
ihonestytruthful You replied to a comment that was made a year ago. A YEAR AGO!!
its never too late to give good reminders!