Bounding Region Inequalities for Solid Rectangular Box in First Octant

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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:
0\leq xyz\leq 6
 
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So, you are claiming that the point
(1000, -1/100, -1/10)​
lies in the box?
 
ok so how about

0\leq x+y+z \leq 6
 
What about the point (100, 200, -300)?

(Oh, and why do you think (1000, -1/100, -1/10) isn't in the box?)
 
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?
 
Winzer said:
Should I just make a contraint for each varible?
That's certainly a reasonable thing to do.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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