OPTIMIZATION: Dimensions that maximize the volume of a cylinder
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- เผยแพร่เมื่อ 27 ม.ค. 2025
- ► My Applications of Derivatives course: www.kristaking...
Optimization problems are an application of derivatives in calculus that allow us to find the local and global extrema of a function, including the local and global minima and the local and global maxima.
In order to find the extrema of a function, you'll need to take the derivative of the function, set it equal to zero, and solve it for the independent variable in order to find critical numbers. These critical numbers represent potential critical points, which are the points at which the function changes direction from increasing to decreasing, or vice versa.
Once you've found the critical numbers, you'll need to use the first derivative test to test each interval between the critical numbers to see where the function is increasing and where it's decreasing. If the function is increasing to the left of the critical number, and decreasing to the right of it, the critical number represents a local maximum, and possibly a global maximum. On the other hand, if the function is decreasing to the left of the critical number and increasing to the right of it, the critical number represents a local minimum, and possibly a global minimum.
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+Перси Давид Thank you!
+CalculusExpert.com It's my pleasure, and an honour to address some words to you. We're on similar professional routes, linked to science and Maths. So, whenever you want to count on a probable fan or support, just feel free to contact me. Best regards.
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I'm happy to help. :)
Wow. Beautifully explained. Thank you.
+lmanderson2012 Thanks!
Beautifully done.
so simple and clear thank you!!
+Rohan Verma You're welcome!
yay!! more optimization!!
Your amazing pls keep making these videos they help so much
Thank you so much, Pauline, I'm so glad the videos have been helping! I'm hoping to start adding more again soon. :)
Ack! This exact problem was on my final YESTERDAY morning. I didn't get it. Oh well. Your other videos have helped me so much.
Thank you for this beatifull work!
+Robson Franklin You're welcome! :)
yes!!! the nicest tutor is back.... and she changed her name lol
what if the perimeter is not given and the volume is given?
Wouldn't the dimensions at the end be = 2x+2y?
Thanks, very clear
+Nadir Yusuf Thanks!
Thanks, good job
+Niels Bos Thank you!
Hi! there is a confusion that the derivative of volume equation "v=(39^2/8Pi)-(x^3/4P)" could be 'v=(39x/4Pi)-(3x^2/(4Pi)^2)' instead of '"v=(39x/4Pi)-(3x^2/4Pi)"?
I've been given a similar problem where a rectangular box is inscribed in a cone z= a-sqrt(x^2+y^2) i have to find the dimensions of the box that maximize it's Vol. Anyone have any ideas on how to solve? Hoping i don't have to change coordinate systems! Great job on the vid as always
Nice
Thank you good stuff
+rejoice dube Thanks!
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I'd never bunk the class if you were my maths teacher...
Math ASMR?