1
$\begingroup$

So I am trying to find the minimum and maximum of the function $f(x,y,z)= x^2 + y^2 - z^2$ on the curve defined by $y^2 + z^2 =1$ and $x=y.$

My work thus far is the following:

$\text{Proof.}$ Let $g=y^2 + z^2 -1=0, h=x-y=0,$ and taking partials, $$f_x = 2x, f_y = 2y, f_z = -2z$$ $$g_x = 0, g_y = 2y, g_z = 2z,$$ $$h_x = 1, h_y = -1, h_z = 0$$ and by definition of the Lagrange multiplier with multiple constraints, we have $\nabla f = \lambda \nabla g + \mu\nabla h$ corresponding to each parameter. Thus, $$\nabla f_x = \lambda g_x + \mu\nabla h_x \implies x = \frac{\mu}{2}$$ $$\nabla f_y = \lambda g_y + \mu\nabla h_y \implies y = \lambda y - \frac{\mu}{2}$$ $$\nabla f_z = \lambda g_z + \mu\nabla h_z \implies z = -\lambda z$$ Now from here I am having issues because I cannot find a solution for $y,z$ Can someone please help me? Thank you.

$\endgroup$
0
$\begingroup$

Assuming your setup is correct up to this point, basically there is a bunch of case work to do. If you're allowed to divide by whatever you want then you can do this:

$$y=\lambda y - \mu/2 \Rightarrow y=\frac{\mu/2}{\lambda-1} \\ z = -\lambda z \Rightarrow \lambda = -1 \\ \Rightarrow y = \frac{-\mu}{4} \\ \Rightarrow x = \frac{-\mu}{4}$$

However we also know $x=\frac{\mu}{2}$, so $\mu=0$, and so you get $z=\pm 1$ so you have the points $(0,0,1)$ and $(0,0,-1)$. You probably don't actually care but the values of the Lagrange multipliers are $\lambda=-1$ and $\mu=0$.

Now what assumptions did you make along the way? You assumed $\lambda \neq 1$ and $z \neq 0$. What happens if you violate one or both of those assumptions?

Incidentally I would say that an easier approach to this particular problem would be to just substitute $x=y$ and do a one-constraint Lagrange problem. Thus you look at $f(y,z)=2y^2-z^2$ subject to $y^2+z^2-1=0$, so $4y=\lambda 2y$ and $-2z=\lambda 2z$, from which you readily conclude that at least one of $y$ or $z$ must be zero (since $\lambda$ cannot be both $2$ and $-1$ at the same time).

$\endgroup$
2
  • $\begingroup$ Thank you, Ian. So we have infinite solutions if we allow for those two cases? Thus, no max or min? $\endgroup$ – shiloh.otis Nov 20 '20 at 23:32
  • $\begingroup$ @shiloh.otis Well, if $\lambda=1$ then the third equation gives $z=0$. Of course $z=0$ itself gives the same thing. Either way, if $z=0$ then the constraints specify two more possible points. Are there admissible values of $\mu$ and $\lambda$ at these points so that all three Lagrange equations are satisfied? $\endgroup$ – Ian Nov 21 '20 at 2:51
0
$\begingroup$

The symmetries in the geometrical arrangement will help somewhat with locating solutions. The constraint surfaces are a circular cylinder with its symmetry axis along the $ \ x-$axis and a plane cutting obliquely through the cylinder. The intersection curve is then an ellipse symmetrical about the $ \ xy-$plane, so we would expect the points at which the extremal values of the function occur to have coordinates $ \ (\pm x \ , \ \pm x \ , \ \pm z) \ = \ (\pm x \ , \ \pm x \ , \ \pm \sqrt{1-x^2}) \ \ . $

enter image description here

For the Lagrange equations, it is sometimes best to bring all or most of the terms to one side in order to factor them; this reduces the risk of overlooking solutions. Here, we would have $$ 2x \ = \ \mu \ \ , \ \ 2y \ = \ \lambda·2y - \mu \ \ \rightarrow \ \ 2y · (1 - \lambda) \ = \ -\mu \ \ , $$ $$ -2z \ = \ \lambda·2z \ \ \rightarrow \ \ -2z · (1 + \lambda) \ = \ 0 \ \ . $$

It is now clearer from the third equation that we have the two cases:

$ \mathbf{z = 0 \ , \ \lambda \neq -1 \ } \ , $ from which it follows immediately that $ \ y^2 \ = \ 1 \ \Rightarrow \ y = x = \pm 1 \ \ , $ for which we have the function value $$ f(\pm 1 \ , \ \pm 1 \ , 0 ) \ \ = \ \ 1^2 \ + \ 1^2 \ - \ 0^2 \ = \ 2 \ \ ; $$

and

$ \mathbf{z \neq 0 \ , \ \lambda = -1 \ } \ , $ for which we obtain $ \ 2y · (1 - [-1]) \ = \ 4y \ = \ -\mu \ = \ -2x \ \ ; $ since the planar constraint requires $ \ y = x \ , $ we must have $ \ x = y = 0 \ \Rightarrow \ z^2 = 1 \ \ , $ leading at once to $ \ f(0 \ , \ 0 \ , \pm 1 ) \ \ = \ \ 0^2 \ + \ 0^2 \ - \ 1 \ = \ -1 \ \ . $

The former case thus gives us the absolute maximum $ \ 2 \ $ and the latter case the absolute minimum $ \ -1 \ $ on the intersection ellipse.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.