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I have not yet had the privilege of studying multivariable calculus, but I have made an educated guess about how to find the minimum or maximum of a function with two variables, for example, $x$ and $y$.

Since, in three dimensions, a minimum or maximum would be represented by a tangent plane with no slope in any direction, could I treat $y$ as a constant and differentiate $z$ with respect to $x$, then treat $x$ as a constant and differentiate with respect to $y$, and find the places where both of these two are equal to zero?

Sorry if this is just a stupid assumption... it may be one of those things that just seems correct but is actually wrong.

Franklin Pezzuti Dyer
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    No, this is pretty close, to find maxima and minima in 3 dimensions (and higher), take a look at this: https://en.wikipedia.org/wiki/Lagrange_multiplier. – 高田航 May 22 '17 at 23:34
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    Differentiating with respect to one variable and keeping the rest constant is called partial differentiation. You are right you would have 2 equations to solve for which values of $x,y$ do both partial derivatives equal to zero and there is also a test to check if it is a minima or maxima like the second derivative test https://en.wikipedia.org/wiki/Second_partial_derivative_test – Ziad Fakhoury May 22 '17 at 23:39
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    @泥九奈加 This method is for finding minima and maxima for functions subject to constraint which is not what OP was asking. – Ziad Fakhoury May 22 '17 at 23:40
  • Maybe this can be helpful,the point need not be a maximum or a minimum for example take $xy$ by example $(0,0)$ is not a extrema but a saddle point. Though you can classify such points. – kingW3 May 22 '17 at 23:41
  • Thank you! This is all very helpful. – Franklin Pezzuti Dyer May 22 '17 at 23:41
  • if $f(x,y)$ is a function of $x$ and $y.$ Holding $y$ constand and differentiating by $x$ is called partial differentiation, sometimes noted as $f_x$ and the partial derivative with respect to $y$ is $f_y$ and second derivatives $f_{xx}, f_{yy}$ and $f_{xy}$. If $f_x = f_y = 0$ and $4f_{xy}^2 > f_{xx}f_{yy}$ you have a saddle. – Doug M May 22 '17 at 23:43
  • This is a great guess! You’ve basically invented partial derivatives on your own. One thing to be careful about is that in multivariable calculus things can go wrong in many more ways. For instance, there are standard examples of functions that have derivatives in the $x$ and $y$ directions at a point, but don’t have a well-defined tangent plane there. Essentially, there’s an explosion of possibilities: in elementary calculus, you only need consistency coming from the right or left; in multivariable calculus, things need to be consistent along all possible paths. – amd May 23 '17 at 00:30

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In fact you got quite a good "guess". When it comes to finding maxima and minima of a function $f:\mathbb{R}^2 \to \mathbb{R}$, you begin by finding the critical points which, as you said, are obtained by solving the system: $$\begin{cases} f'_x(x_0, y_0) = 0 \\ f'_y(x_0, y_0) = 0\end{cases}$$ The solution(s), $(x_0, y_0)$, is either going to be maxima, minima or saddlepoint. You determine which one it is by studying the Hessian matrix evaluated at that point.

I'm sure you will get to it when you start doing multivariable calculus. Great guess, however!

BB3C
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That's actually a necessary condition for well-behaved functions, but it's not sufficient.

Classical counterexample:

$$ f(x,y) = x^2 - y^2 $$

satisfies your condition in $(0,0)$ but it's not a maximum (nor minimum), it's a saddle point.

saddle point from Wikipedia, author: Nicoguaro

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My suggestions are:

  1. Check second-order derivatives as well to make sure the point you get is a local maximum and minimum
  2. Then also check the boundary points (if there is any), because values at boundary could be larger than local maximum or smaller than local minimum
  3. Check the function behavior as it goes to infinity - you'll probably find values there larger than local maximum or smaller than local minimum
  4. Also pay attention to where derivatives do not exist

I also put a similar answer in this post in case you are interested in it.

Jay Zha
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