# Why do we need both partial derivatives for $f(x, y)$ to guarantee that we can find min & max points?

Given for example a function of 2 variables $z = f(x,y) = x^2 + xy + y^2$, I believe there are infinitely many directions on the $xy$ plane in which we could take partial derivatives (this could already be wrong...).

If we differentiate $z$ w.r.t $x$, we get $f_x = 2x+y$. If we set this partial derivative equal to zero, we get infinitely many solutions; namely it seems that for any value of $y$, we could find a value of $x$ such that at the point $(x,y)$ the slope in the direction of the $x$ axis would be zero.

However, only one of these infinitely many points is actually the minimum of the function.

But if we take the other partial derivative $f_y = 2y + x$ and set both of them equal to zero as a system of equations and solve that system, we will get our minimum point at $(0,0)$.

Why is that? Why is it that we need both partial derivatives and not fewer, or especially more?

EDIT: Just to make it clear, I'm wondering both why two PD's are the minimum, and also why they are guaranteed to be enough (as in, why don't we need 20 PD's?).

• This question is unclear. IF the point is a maximum/minimum, then the partial derivatives (any) are $0$. The converse does not hold. If this question asked "Why are two PD being equal to $0$ throughout a domain sufficient to prove that the function is constant", then the question could admit a satisfying answer. – Aloizio Macedo Jan 5 '16 at 18:32
• Note that, under mild assumptions, if both partial derivatives are $0$ at $x$, then all directional derivatives at $x$ must be $0$. – littleO Jan 5 '16 at 18:43
• @AloizioMacedo, the question is unclear partly because of the confusion that prompted the question in the first place. By voting to close those kinds of questions you create a catch 22 whereby a whole subset of confused people asking legitimate clarifications are unable to do so because they don't understand the subject well enough to properly express their confusion. – jeremy radcliff Jan 5 '16 at 18:45
• @littleO, that's interesting and now makes sense to me in the context of Trevor's answer, thanks. – jeremy radcliff Jan 5 '16 at 18:47
• @jeremyradcliff I understand your point, but I don't think this is the case. There is a difference between a confusion which actually asks something (like this one math.stackexchange.com/questions/1601095/…), so one person can answer properly and the confused asker can verify whether what he was confused about is/is not true, and your question. Your question is "Why is it that we need both partial derivatives and not fewer, or especially more?" in order to determine maxima/minima, and as it is(...) – Aloizio Macedo Jan 5 '16 at 18:56

The analog of finding the tangent line for a function $f(x)$ is finding the tangent plane of $f(x,y)$. So just how one might find the max/min of $f(x)$ by finding where the tangent line is flat, one can find the max/min of $f(x,y)$ by finding points where the tangent plane is flat (consider a "hill" in 3-d space to convince yourself of this).

To determine a plane in 3-d space, one needs at least two (linearly independent vectors) vectors. Therefore checking that vectors are flat in the $x$ and $y$ directions is enough to uniquely determine the entire tangent plane. If we only had one vector, there would be infinitely many planes which could go through it.

I believe one could take the directional derivatives of any two linearly independent unit vectors and check that they are zero to determine if the tangent plane is flat. However the $x$ and $y$ directions tend to be the most natural.

• This makes perfect sense, it's exactly the kind of insight i was looking for. Thank you. – jeremy radcliff Jan 5 '16 at 18:42
• Glad I could help :) – Trevor Norton Jan 5 '16 at 18:47

The following picture shows why we need two directions: You see that the only minimum point is the red dot. But if we only find the derivative along, for example, $y$-direction, we get the whole bottom curve, which are minimums along the cross-sections parallel to the $y$-axis.

The reason why we don't need more directions is because the directional derivative can be represented by the derivative in $x$ and $y$ directions. So if these two are $0$'s, the other derivatives should also be $0$'s.

• Your graph actually makes it easier for me to understand what's going on, thank you. – jeremy radcliff Jan 5 '16 at 18:49
• @jeremyradcliff: Glad to know that. – KittyL Jan 5 '16 at 18:54