Great way to explain, and yes, he didn't actually answer the question but matters that youe've learn how to solve this kind of problems, that is enough :)
@storaman12 Yeah, the amount of cardboard used isn't in square units. Multiplying by the thickness of the cardboard gives it volume. Although the method is the same, the terminology used is incorrect.
He suggests that being local minimum isn't enough for this function, and we need a global min. However since there's only one critical point for all x>0 and y>0, and the 2nd derivative test shows it's a minimum, then it's not possible for the endpoints to do anything else other than grow exponentially. I get that it's sometimes/often necessary to check boundary values/behavior, but is that really applicable in this case?
very comprehensive,
thank you for the detailed explanations
dimensions are 1.5 height, 2 width and 1 lenght
I love you Joel
Great way to explain, and yes, he didn't actually answer the question but matters that youe've learn how to solve this kind of problems, that is enough :)
@storaman12 Yeah, the amount of cardboard used isn't in square units. Multiplying by the thickness of the cardboard gives it volume. Although the method is the same, the terminology used is incorrect.
He suggests that being local minimum isn't enough for this function, and we need a global min. However since there's only one critical point for all x>0 and y>0, and the 2nd derivative test shows it's a minimum, then it's not possible for the endpoints to do anything else other than grow exponentially. I get that it's sometimes/often necessary to check boundary values/behavior, but is that really applicable in this case?
Thanks, big help. and i feel like i am getting a bit of an MIT education
I suppose the volume is not given by this formula when one allows for thickness...
Wrong;, x.y=area defined, then defined as thickness; makes non sense.
Can i use Lagrange Multiplier for this exercise ? If can not, why ? there is a constraint.
there is not one critical point, (2,-1) is also the critical point
Dude, he mentioned-no negatives.
this can be very easily solved by amgm
9:55 What about -2?
5:33
oh, sorry. x have to be greater than 0!