Optimization problem

[dbag]

Hello

Problem statement:
plate of lenght 1000mm and width 150mm is bent so it forms a covered channel(rectangular cross-section). Determine the height of the channel so that volume is maximized




Solution:
The constraint to me is clearly the amount of material available but what is not so clear is how would i go about solving it?

1. Function to maximize is volume, so the lenght of that channel would have to be either 1000mm or 150mm depending on which way you bend.
2. The constraint gives me total area of plate to waste, how does that help me? Volume is area*height.

Thanks

 

[Deleted member 4993]

Hello

Problem statement:
plate of lenght 1000mm and width 150mm is bent so it forms a covered channel(rectangular cross-section). Determine the height of the channel so that volume is maximized


View attachment 14172

Solution:
The constraint to me is clearly the amount of material available but what is not so clear is how would i go about solving it?

1. Function to maximize is volume, so the lenght of that channel would have to be either 1000mm or 150mm depending on which way you bend.
2. The constraint gives me total area of plate to waste, how does that help me? Volume is area*height.

Thanks

So

either the volume of the channel is = x * y * 150 & in that case 2x + 2y = 1000

or

the volume of the channel is = x * y * 1000 & in that case 2x + 2y = 150

continue.....

 

[dbag]

square 2500x2500x150 volume is 937,5l

 

[Steven G]

What does your reply mean? Are you saying that when you square 2500x2500x150 you get 937,5l? That is not true!

 

[dbag]

oh no very sorry.

I bend the plate so that the lenght of circumference is 1000mm, then the largest area
the volume of the channel is = x * y * 150 & in that case 2x + 2y = 1000
Volume gets maximized y(500-y)*150 gets derived and set to zero. and out of it i get a lenght of side y 250mm
Larges area from a square, then the volume is 150mmx250mmx250mm and that's 9,375L