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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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    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.2012-02-19

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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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    Yes, that was my guess. Please tell me if everything is clear now,2012-02-20