This was so helpful, thank you. I honestly thought I was dumb for not understanding the way my teacher taught it. But the way you explained it clicked for me. ❤🎉
I know it will likely be too late by the time anyone sees this, but I’m using this for a project and my original box was 17cm in length and 7.5cm in height and had a surface area of 1088cm cubed. Using my surface area I was supposed to maximize the volume and by doing so it grew on both sides to 19cm and 9.5cm and I haven’t found any problems in my math, so what did I do wrong if anything?
but there is only 1200 cm2 available ....if it was a closed box, it would take some of the material to cover the top ...thus the material remained for the sides and the base would be less and thus the maximum volume would be smaller...
@@shazinhamasaeed7037 ^^^the alteration to this process would be changing the constraint equation to be 2x^2+4xh=1200 cm2, since the top is identical to the base we're given. Once you make that change, you should be able to follow the same steps shown, and then the new value of h you find will account for the fact you needed to use more surface area on the new side.
The way you explain this problem step by step really helped me understand these types of problems. Thank you so much, truly.
You’re welcome. Glad it helped.
This is single-handedly the best video I have ever seen on this problem. I love you so much
Thank you.
Throwing out a thank you signed at 2:30 am on a monday- homework is due at 8:00 and you're a life saver.
You’re welcome. Glad it helped.
This was so helpful, thank you. I honestly thought I was dumb for not understanding the way my teacher taught it. But the way you explained it clicked for me. ❤🎉
You’re welcome. Thanks for the comment. Glad it helped.
lifesaver!!! i struggled for hours before coming across this video. thanks :)
You’re welcome.
Actually videos that are related to the difficulty in my class great job
Thank you.
Simple and to the point, very good
Thank you.
Thank you so much! Helped with a difficult homework problem
You’re welcome. Glad it helped.
Super helpful. Thank you!
You’re welcome. Glad it helped.
Thank you Steve, this was very helpful!
You’re welcome. Glad it helped.
Is there a question on optimisation where there are three different variables, instead of a square base (having just two variables)?
Amazing Video! Very helpful for my class!
Thank you
thank you so much this helped me greatly
You’re welcome. Glad it helped.
Thank you, this was very helpful
You’re welcome. Glad it helped.
u is a real G appreciated it G
Thank you.
Super helpful! Thanks man
You’re welcome. Thank you for the comment.
Thank you this helped me a lot!
You’re welcome.
hey can you do a question with a rectangle instead of a square?
Thank you for this explanation 👍
You’re welcome.
super helpful, thank you so much
You’re welcome. Glad it helped.
I know it will likely be too late by the time anyone sees this, but I’m using this for a project and my original box was 17cm in length and 7.5cm in height and had a surface area of 1088cm cubed. Using my surface area I was supposed to maximize the volume and by doing so it grew on both sides to 19cm and 9.5cm and I haven’t found any problems in my math, so what did I do wrong if anything?
Thank you!
You’re welcome.
Wonderful job
Thank you.
what happen if you get a min answer and the max is negative
thx homie
You’re welcome.
useful
ty
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How did you go from v to v1 at 5 minutes. I don't get how the -1/4 changed to -3/4.
Same i was wondering the same thing
Thank you (:
You’re welcome.
god bless 🙏🏽
Thank you.
Nice one
Thank you.
slow: 👍, explained:👍, crowded writing: 👎
I thought it was fine and the writing made it easy to follow
TY SMMM!!!
You’re welcome.
Thankyouuu
You’re welcome.
If it was closed would it be different
Yes. You would have to factor in the top.
Steve Crow but if it’s for volume wouldn’t the volume be the same with with or without a top or lid
but there is only 1200 cm2 available ....if it was a closed box, it would take some of the material to cover the top ...thus the material remained for the sides and the base would be less and thus the maximum volume would be smaller...
@@shazinhamasaeed7037 ^^^the alteration to this process would be changing the constraint equation to be 2x^2+4xh=1200 cm2, since the top is identical to the base we're given. Once you make that change, you should be able to follow the same steps shown, and then the new value of h you find will account for the fact you needed to use more surface area on the new side.
R u from texas
ruclips.net/video/paVOEi7cYrA/видео.html (Mec.English and French )👍💐