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What is the graph of$|x|+2|y|+3|z|\leq 1$ ? And determine its geometrical figure. I know that the graph will be 3D. If I assume that $x \geq 0$,$y \geq 0$,$z \geq 0$. So let $x'^2 = x \geq 0$,$y'^2 = y \geq 0$,$y'^2 = z \geq 0$. Then above inequality can be written as $x+2y+3z\leq 1$ and $x'^2+2y'^2+3z'^2\leq 1 \implies \frac{x'^2}{1}+\frac{y'^2}{1\over2}+\frac{z'^2}{1\over3}\leq 1$ which is ellipsoid and its interior. Please help me.

EM4
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Sangam Academy
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  • It may help to plug it into desmos: https://www.desmos.com/calculator/29urwl3y1z – TomKern Jul 04 '21 at 03:43
  • It's the graph of $|x|+|y|+|z|\le1$ compressed along the $y$-axis by a factor of $2$ and along the $z$-axis by a factor of $3$. Do you know what the graph of $|x|+|y|\le1$ is in 2D? – anon Jul 04 '21 at 03:48
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    the condition $x+2y+3z=1$ is a plane – janmarqz Jul 04 '21 at 03:48
  • @runway45, Yes it is a diamond like figure. – Sangam Academy Jul 04 '21 at 03:52
  • octahedron, one of the faces of which is the plane $x+2y+3z=1$ intersected with the first octant. It becomes an ellipsoid with respect to the new variables $x',y',z'$ that you introduce, but these new and different variables go with a different coordinate system, not the coordinate system that goes with the original variables $x,y,z$ – Mirko Jul 04 '21 at 03:59
  • the eight combinations $x+2y+3z=1$, $-x+2y+3z=1$, $x-2y+3z=1$, ...etc are planes that bound a diamond shape solid – janmarqz Jul 04 '21 at 04:00
  • janmarqz is correct - and since the point (0,0,0) is inside the body, it is its volume. – Moti Jul 04 '21 at 05:53

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