Consider the prolem where we need to both minimize and maximize $f(x,y) = x^2+y^2$ subject to $x = y+1$. (So $g(x,y) = x-y=1.)$

Setting up the Lagrange Multiplier equation $\nabla f(x,y) = \lambda \nabla g(x,y)$, we obtain $$ \langle 2x,2y \rangle = \lambda \langle 1, -1\rangle $$

which implies that $2x=\lambda$ and $2y=-\lambda$

Now, if $x = 0$, then $\lambda = 0$, which would imply that $y=0$, but this does not satisfy $g(x,y) = 1$. In fact, solving $2x=\lambda$ and $2y=-\lambda$ for $x = \lambda/2$ and $2y=-\lambda$ for $y = -\lambda/2$ and substituting into $g(x,y) = x-y = 1$ gives us that $$ g(x,y)=x-y = 1 \\ \implies \frac{\lambda}{2} - \left(-\frac{\lambda}{2}\right) = 1 \\ \implies \lambda = 1 \\ \implies x = \frac{1}{2}\, \text{and}\, y = -\frac{1}{2} $$

So, our only critical point is $\displaystyle \left(\frac{1}{2}, \frac{-1}{2} \right)$, which in the context of the problem cannot be both the max and the min.

My Question is: I was told that in this problem, $f$ attains its max, but its min is not attained, and I was told that in order to explain this, I need to explain the "boundary behaviour" of $f$ - i.e., what happens as we go to $\infty$ along the line $x = y+1$. However, I don't understand what is meant by that - could somebody please explain it to me?

I figured drawing a picture would be helpful, so I'm including it here: enter image description here Essentially, what I did was draw a bunch of level curves of $f$ for various constants $k$, and draw the line $g$ on the same graph, but I don't understand what is happening as we go to $\infty$ along $g$. I have to teach this to a bunch of undergrads in a few hours, and I'd like to understand what I'm trying to teach them, so if you could kindly explain it to me, I would appreciate it more than you know! Thank you.


For every $M\in\mathbb R$, there exists a point $(x,y)$ on the line $y=x+1$ such that $f(x,y)>M$. In particular, one such point would be $x=\max\{1,M\}, y=\max\{1,M\}+1$.

Therefore, the function is not bounded on the line (and has no maximum).

Another way of saying the same thing would be to say this:

If we parametrize the line as $(t,1+t)$, with $t\in\mathbb R$, then $$\lim_{t\to\infty}f(t,1+t) = \infty$$

  • $\begingroup$ let me get this straight: for every $M \in \mathbb{R}$, $\exists (x, x+1)$ such that $f(x, x+1) = (x)^{2} + (x+1)^{2} > M$? $\endgroup$ – ALannister Nov 21 '17 at 9:38
  • $\begingroup$ @ALannister Exactly. For example, for $M>1$, you have $f(M,M+1)>M$. $\endgroup$ – 5xum Nov 21 '17 at 9:41
  • $\begingroup$ thanks a lot! I think you might have just pulled my bum out of the fire. $\endgroup$ – ALannister Nov 21 '17 at 9:41
  • $\begingroup$ @ALannister To be honest, you did most of the work yourself with finding the minimum. Showing there is no maximum is actually the easier part, once you get the idea. $\endgroup$ – 5xum Nov 21 '17 at 9:42
  • 2
    $\begingroup$ @ALannister Well, you know that the function has a minimum (because it is bounded from below by $0$, and it's value is large outside some bounded - compact - set). You also know that the discovered point is either a minimum or a maximum, therefore it has to be the minimum. So you don't need second derivatives for that. $\endgroup$ – 5xum Nov 21 '17 at 9:46

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