Showing that $\Omega=\{x\in\mathbb{R}^3 : 4x_1^2+x_2^2=16 \ \ \text{and} \ \ 2x_2+3x_3=25\}$ is bounded

Question

I am trying to show that the set $$\Omega=\{x\in\mathbb{R}^3 : 4x_1^2+x_2^2=16 \ \ \text{and} \ \ 2x_2+3x_3=25\}$$ is bounded. My attempts have been to find $$ \|x\|=\sqrt{x_1^2+x_2^2+x_3^2}\leq M, \ \forall x\in\Omega. \tag{1}$$ I have tried to substitute the conditions $4x_1^2+x_2^2=16\iff x_2^2=16-4x_1^2$ and $2x_2+3x_3=25\iff x_3^2=1/9(25-2x_2)^2$ into $(1)$, which did not seem to help find an $M>0$.

Any suggestions are welcomed.

Answer 1

If $(x,y,z)\in\Omega$, then $4x^2+y^2=16$, and therefore, $|y|\leqslant4$ and $|x|\leqslant2$. Since $|y|\leqslant4$ and $2y+3z=25$, you have $z=\frac{25-2y}3$, and therefore$$|z|=\left|\frac{25-2y}3\right|\leqslant\frac{25+2\times4}3=11.$$So$$\|(x,y,z)\|\leqslant\sqrt{2^2+4^2+11^2}<12.$$

Answer 2

You can proceed using rough inequalities:

$(3x_3)^2=(25-2{x_2}^2)^2\implies 9{x_3}^2=25^2-100x_2+4{x_2}^2$

Then use $|x|<1+x^2\quad$ to get $\quad 9{x_3}^2-104{x_2}^2\le 725$

In fact we can just use $x^2\le kx^2$ for any $k\ge 1$ therefore by adding $105$ times the first equation we get:

$${x_1}^2+{x_2}^2+{x_3}^2\le 105(4{x_1}^2+{x_2}^2)+(9{x_3}^2-104{x_2}^2)\le \overbrace{16\times 105+725}^\text{some cst}$$

There are probably some clever ways to establish a more precise inequality, but I wanted to show that to only prove boundedness, you can often proceed very roughly.