Maximize Volume of an Open Top Box (Optimization) | Calculus 1 Exercises

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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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Комментарии
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Follow along with my AP Calculus FRQ Advent Calendar!

WrathofMath
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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

SilentPenguin
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Thank you, very helpful. And having Nicolas Cage do do your voice over was cool.

leighahrens
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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.

larry
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Keep it up man.... excellent work 👌🏻👌🏻

SaurabhMishra
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very helpful, having a AP calc test coming up soon

ooh_netiyiy
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Interesting fact: the volume is 108 in cu. In. Just like the surface area is 108 sq. In.

larry
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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 ?

christaylor
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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?

anthonychuah