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Let $f : \mathbb{R}^3 \to \mathbb{R}, f(x, y, z) = x + y + z$ with constraints $$x + z = 1 \text{ and } x^2 + y^2 = 4 $$

Now it says "The constraint set is the intersection of a cylinder [I understand this, with radius $4$ and $z$ arbitrary] and a plane [Obv $y$ is arbitrary here but how can I interpret $1$ here?]. Since the plane does not intersect the cylinder perpendicularly [because $z \neq 0$?], the intersection is an ellipse and, therefore, compact." Why is is important that the intersection is an ellipse? Wouldn't this be true if the interection was a circle? For $x=0$ the intersection would be an infinite line, wouldn't this be 'worse'?

user1049882
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3 Answers3

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Actually the radius of the cylinder is $2$, not $4$.

Any equation $ax + by + cz = d$ with $a,b,c$ not all $0$ is the equation of a plane.

It is indeed not important that the intersection is not a circle.

Robert Israel
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The $x+z=1$ is a plane, because you can think of it as a line on the $xz$ plane $z = 1 - x$ which is then propagated along the $y$ axis.

"The intersection is a circle if $x=0$" is simply an irrelevant claim. $x$ is not 0, it appears in the equation of the surface you're intersecting. The plane is a constant object, with a unique formula. There are no "parameters" here of which you're free to chose the value. If $x=0$ then $z=1$, and $y$ is arbitrary.

The fact the the intersection is an ellipse is indeed irrelevant for claiming it's compact, but may be relevant in the following text which you omitted.

The fact that the intersection is compact is non-trivial a-priori, since you're intersecting two non-compact surfaces. The reason it is compact is because $z$ is bounded in the plane eq. as long as $|x|<2$ (which is a constraint of the cylinder equation), so you don't really need the actual geometric shape for that.

Amit
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  • Thank you for your answer!! Is this a visual representation https://math.stackexchange.com/questions/1870146/cylinder-defined-on-3d-coordinate-plane? – user1049882 Jul 02 '23 at 17:02
  • Exactly (referring to one of the images in the answers) – Amit Jul 02 '23 at 17:09
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How can I interpret 1?

If $x=0$, then $z=1$, similarly if $z=0$, then $x=1$. So, you can interpret 1 as the intersection on the other axis. It's like a line in the $zx$ plane, with the only difference being confined in a space of 3 dimensions.

The intersection is not perpendicular because $z\neq0$?

Actually, because $z\neq k$, for $k\in\mathbb{R}$, i.e. the $z$ value is not constant. If it had been constant, it might have been a plane parallel to the $xy$ plane (e.g. z=3), and thus the intersection would've been at 90 degrees, clearly.

Why is it important that the intersection is an ellipse?

Based on what you have written, it's not relevant, it could've been anything. One reason why it might be useful is in the context of optimization. As a matter of fact, by minimizing/maximizing $f$ in a compact set, you'll have sufficient guarantees of finding minimizers/maximizers.

Sandro
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  • Thank you for your answer! Yes, we need compactness for finding minimizers/maximizers. But why do we need an ellipse? Do we lose compactness in case of a circle? – user1049882 Jul 02 '23 at 17:28
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    You don't need an ellipse, but any compact set is fine. A circle is compact. – Sandro Jul 02 '23 at 17:49