Find the dimensions of a rectangular box, open at the top, with a volume of 32 cubic feet, using the least amount of material.
I'm not really sure even how to start this problem. Any help?
Are you familiar with Lagrange multipliers?.
We must minimize
subject to the constraint
If we let and
then,
and
If follows that at any point on the surface
xyz=32, since x, y, and z are all non zero on this surface. So, we must have
So, we get the 3 equations:
These can be rewritten like:
...[1]
....[2]
....[3]
From [1] and [2] we get
From [2] and [3] we get
Sub these into the constraint and get:
Now, I will leave finding the values up to you. Not much left to do.
You can also do this with partial derivatives.
Hello, ikickyouindanuts!
You need to be familiar with partial derivatives . . .Find the dimensions of an open-top rectangular box with a volume of 32 ft³,
using the least amount of material.Code:*-----------* /| /| / | / | / | / |H *-----------* | | | | | | * H | | / | | /W | |/ *-----------* L
The volume is 32 ft³: .
The base has area: .
The front and back has area: .
The sides have area: .
The total area is: .
Substitute [1] into [2]: .
. . and we have: .
Take partial dereivatives and equate to zero:
. .
Equate [3] and [4]: .
Substitute into [3]: .
Substitute into [1]: .
. . Therefore: .