
#1
Mar312, 03:09 PM

P: 19

1. The problem statement, all variables and given/known data
Show that [itex]D = { (x,y,z) \in \mathbb{R}^{3}  7x^2+2y^2 \leq 6, x^3+y \leq z \leq x^2y+5y^3}[/itex] is bounded. 2. Relevant equations Definition of bounded:[itex]D \subseteq \mathbb{R}^{n}[/itex] is called bounded if there exists a M > 0 such that [itex]D \subseteq \{x \in \mathbb{R}^{n}  x \leq M\}[/itex] 3. The attempt at a solution I have to find a M such that [itex]D \subseteq \{(x,y,z) \in \mathbb{R}^{3}  x^2 + y^2 + z^2 \leq M\}[/itex]. I thought of just picking a very high M, say 999999. But how do I show it works? 



#2
Mar312, 03:12 PM

Mentor
P: 4,499

It's often easier to show that each coordinate is bounded, say x,y and z are all smaller than 10 maybe. Then (x,y,z) < (10,10,10) = M




#3
Mar312, 05:16 PM

P: 19

So if I say something like: [itex]7x^2+2y^2 \leq 6 \Rightarrow y^2 \leq 3 \lt 4 \Rightarrow y \lt 2[/itex]
and [itex]7x^2+2y^2 \leq 6 \Rightarrow x^2 \leq \frac{6}{7} \lt 1 \Rightarrow x \lt 1[/itex] and [itex]z \leq x^2y+5y^3 \Rightarrow z \lt (2+5*8)=42[/itex] So choose M = 4+1+42^2 = 1769. And this M will do. 



#4
Mar312, 05:33 PM

Mentor
P: 4,499

Show set (which is a subset of R^n) is bounded
Careful, M=(1,2,42) is the square root of the number you put up. As long as M is larger than 1 that won't matter, but if the norm happens to be smaller than 1 failing to take the square root can give a value of M that doesn't work



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