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In my math lectures, I learnt that an extremum of a function $f:\mathbb R^2\to \mathbb R$ requires $\mathrm{grad}(f)=0$. So if $f$ was $f(x_1, x_2)$ that means $(∂f/∂x_1, ∂f/∂x_2) \cdot (x_1, x_2)^{\top}=0$. (Sorry for my poor Tex skills, am working to improve those).

No I came across the following in an economics lecture and I can't figure out if what they do is correct:

To find an extremum in $\pi=p*f(x_1, x_2) - w_1x_1 - w_2x_2$ they claimed it was sufficient to find a point satisfying $\frac{∂\pi}{∂x_1}=0$ and $\frac{∂\pi}{∂x_2}=0$

In contrast, the normal gradient approach would yield $\frac{∂\pi}{∂x_1}x_1 + \frac{∂\pi}{∂x_2}x_2=0$ as the precondition for an extremum.

It's pretty clear that $\frac{∂\pi}{∂x_1}=0$ and $\frac{∂\pi}{∂x_2}=0$ implies $\frac{∂\pi}{∂x_1}x_1 + \frac{∂\pi}{∂x_2}x_2=0$, but not the other way round. Because of that, I'd say that the approach used in the economics lecture might not find all interesting points for an extremum.

Is that correct? Or is there anything I have overlooked?

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I've edited your question so that you can see how to TeX a little better. In particular, the transpose is written using \top, which gives $^{\top}$ instead of $^T$. And I've put $x1$'s as $x_1$'s too. If you want to see the edits, where it is written "edited ** mins ago" above my name, just click the "** mins ago" in blue and you'll see my edit. –  Patrick Da Silva Feb 19 '12 at 17:16
@PatrickDaSilva: Thanks for your tips. I'm used to the way SE works from Stackoverflow.com :-) Are you guys using a graphical editor for Text typing all that by hand? Might help the novice. (You might get used to it, pretty much like programming languages I guess). –  Johannes Rudolph Feb 19 '12 at 17:18
All what you see in the editing link is hand-typed. There's no program that gives you any shortcut to what you see there. At least not one that I'm using. –  Patrick Da Silva Feb 19 '12 at 17:44

1 Answer 1

up vote 1 down vote accepted

The correct approach is the following: the necessary condition for a point $(x_1,x_2)$ to be an extremum of $C^1$ (i.e. continuous differentiable) function $f$ is: $$ \nabla f(x_1,x_2) = 0 $$ i.e. both derivatives are zero: $\partial_1f(x_1,x_2) = 0$ and $\partial_2 f(x_1,x_2) = 0$. Due to your notation I guess that you have misunderstood the formula $$ \left(\frac{\partial f}{\partial x_1},\frac{\partial f}{\partial x_2}\right)(x_1,x_2) = 0 $$ as a product while it is an evaluation at a point $(x_1,x_2)$ instead.

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Thanks for your answer (and the latex hints!). I'm german, so I wasn't sure if $grad(f)=0$ was called necessary precondition. So are you saying that $\nabla f(x_1,x_2) = 0$ <=> (exactly when?) $\partial_1f(x_1,x_2) = 0$ and $\partial_2 f(x_1,x_2) = 0$ –  Johannes Rudolph Feb 19 '12 at 17:14
I think my misunderstanding was based on some notational confusion: $\nabla f(x_1,x_2) = 0 = (0, 0)$. Thats why I assumed there must be dot product involved, so I get from $R^2$ into $R$, because I did interpret 0 as a scalar. –  Johannes Rudolph Feb 20 '12 at 7:00
Yes, that was my guess. Please tell me if everything is clear now, –  Ilya Feb 20 '12 at 8:03

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