Maximize Volume of an Open Top Box (Optimization) | Calculus 1 Exercises
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- čas přidán 27. 08. 2024
- We solve a common type of optimization problem where we are asked to find the dimensions that maximize the volume of an open top box with a square base and a fixed surface area. To do this we begin with a volume equation then use the surface area restriction to rewrite the volume in terms of a single variable, at which point we can use our usual strategy of taking the first derivative, finding critical points, then using the first or second derivative test to classify these critical points. #Calculus1 #apcalculus
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The way you explained how to solve this is seriously awesome. I was struggling with this type of problem and your teaching made it a cake walk. Thanks
Thanks a lot - so glad it helped! I've got several other optimization problems you may find helpful in my Calc 1 exercises playlist: czcams.com/play/PLztBpqftvzxUEqGGgvL3EuIQUNcAdmVhx.html
Let me know if you ever have any questions!
Thank you, very helpful. And having Nicolas Cage do do your voice over was cool.
Glad it was helpful! Do I sound like Nicolas Cage? I don't think I've had that comparison before haha
If the box were closed the dimensions would be 3sqrt2 in. X3sqrt2in.X3sqrt2in. In addition to working out the math the same way, this makes sense because a perfect cube has the maximum volume for a rectangular box with all closed sides.
i encountered a similar quirk with a capsule problem (min surface area for a given volume or something), and all it wanted was the radius of the hemispheres but working out thenproblem further the min shape was just a sphere, capsule formed around a disk of height 0
Wrath of Math is its own classic! 😀
I'd hate to think of any of my videos as classics haha, but some of them at this point came out like 6 years ago, I'm getting old! 😅
Keep it up man.... excellent work 👌🏻👌🏻
Thank you!
very helpful, having a AP calc test coming up soon
Glad to help, assuming you're talking about AP Calc - good luck on Monday!
can you do the harder extension to this problem where we're not guaranteed that the base is square? so "maximize the volume of a rectangular-based open-top box". I think the proof that the base is square is intuitive, but could you do it set up like a calculus problem with max (xyh) w.r.t. x, y holding h constant?
Interesting fact: the volume is 108 in cu. In. Just like the surface area is 108 sq. In.
Would the surface area equation not be = to 108=x^2 +5xh due to there being a 5th square face on the bottom of the cube ?
Thanks for watching and no - the four faces are not squares, they are rectangles x by h. So we have four of those, so 4xh. the square base you refer to is where the x^2 comes from. Hope that helps!
Thanks eyy
Thanks for watching!