Bounding Region Inequalities for Solid Rectangular Box in First Octant

  • Thread starter Winzer
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In summary, the conversation is about writing inequalities to describe a solid rectangular box in the first octant bounded by the planes x=1, y=2, and z=3. The attempt at a solution is to use the volume of the box as the constraint, resulting in the inequality 0≤xyz≤6. However, there is some discussion about whether specific points, such as (1000, -1/100, -1/10) and (100, 200, -300), are actually in the box. The suggestion is made to make a constraint for each variable.
  • #1
Winzer
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Homework Statement


Ok I just wanted to make sure of this one.
Write inequalities to describe the region:
The solid rectangular box in the first octant bounded by the pane x=1 y=2 z=3.

The Attempt at a Solution


I thought of it as the volume of the box bounded boy the planes so:
[tex]0\leq xyz\leq 6[/tex]
 
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  • #2
So, you are claiming that the point
(1000, -1/100, -1/10)​
lies in the box?
 
  • #3
ok so how about

[tex]0\leq x+y+z \leq 6[/tex]
 
  • #4
What about the point (100, 200, -300)?

(Oh, and why do you think (1000, -1/100, -1/10) isn't in the box?)
 
  • #5
Hurkyl said:
What about the point (100, 200, -300)?

(Oh, and why do you think (1000, -1/100, -1/10) isn't in the box?)

They are all in the box. Should I just make a contraint for each varible?
 
  • #6
Winzer said:
Should I just make a contraint for each varible?
That's certainly a reasonable thing to do.
 

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