Multivariable Optimization Problem

  • #1
I have two questions.

A) Show the parallelipided with fixed surface area and maximum volume is a cube.

I've already proven that we can narrow down the proof to a box. So, basically, I'm really lost on how do prove that a cube is the box with a fixed surface area and maximum volume.

B) We might not have covered how to do part B yet, so i'll create a new topic if I still don't understand after tomorrow's lecture.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
956
How is a "box" different from a parallelpiped?

Call the lengths of the sides of your parallelpiped x, y, and z.

The volume is V= xyz.

The surface area is 2xy+ 2xz+ 2yz= A (a constant).

Now use the "Lagrange multiplier" method.

In order that V= xyz be a minimum (or maximum!) on the surface U=2xy+2xz+ 2yz- A=0, the two gradient vectors, grad V= <yz, xz, xy> and grad U= <2y+ 2z,2x+ 2z, 2x+ 2y> must be parallel. That is we must have <yz, xz, xy>= some multiple of <2y+2z, 2x+ 2z, 2x+ 2y> so that yz= &lambda;(2y+ 2z), xz= &lambda;(2x+ 2z), and
xy= &lambda;<2x+ 2y>. Eliminate &lambda; from tose equations and see what happens.
 
  • #3
arildno
Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,970
132
A box is characterized with right angles, whereas a parallellepiped need not be subject to this constraint.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Ah, right. Thanks.
 

Related Threads on Multivariable Optimization Problem

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
4
Views
1K
Replies
2
Views
2K
Replies
6
Views
905
Replies
4
Views
1K
  • Last Post
Replies
11
Views
4K
Top