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A question from a past examination paper reads as follows:

Use the method of Lagrange multipliers to find the extreme points of the function $$f(x,y,z) = x^{2}+y^{2}+z^{2}$$ subject to the constraint $$(x-3)^{2}+(y-2)^{2}+(z-1)^{2}=1$$

I have managed to get to the point where I am left with a system of equations from solving $\frac{\partial f}{\partial x}$ , $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial z}$, but unfortunately I can't seem to solve for $\lambda$.

I have tried multiple approaches to no avail and I am still left with $$ \begin{cases} x = \lambda (x-3) \\ y = \lambda (y-2) \\ z = \lambda (z-1) \\ \end{cases} $$ Taking a step back to look at this problem it seems really simple and I feel I should be able to crack out an answer effortlessly but alas, the easiest of problems seem to be the most difficult.

I would appreciate it if someone could tell me how I could approach this.

Cheers.

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From \begin{align*} x &= \lambda(x-3)\\ y &= \lambda(y-2) \\ z &= \lambda(z-1) \end{align*} we see that $z \neq 1$ and $\lambda = z/(z-1)$ whence $y(z-1) = z(y-2)$ which gives $y = 2z$. Likewise $x(z-1) = z(x-3)$ or $x = 3z$. Plugging these into the constraint equation $$(x-3)^2 + (y-2)^2 + (z-1)^2 = 1$$ gives $14(z-1)^2 = 1$ or $z = 1 \pm \frac{1}{\sqrt{14}}$. From this you get $x$ and $y$ easily.

As has been pointed out in another answer, the more natural approach here is a geometric one: draw a line from the origin to the point $(3,2,1)$ and find the point on the line which lies on the sphere of radius $1$ centered at $(3,2,1)$.

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  • $\begingroup$ Thank you very much, this explains a lot. And although the more natural approach may indeed be to look at the problem from a geometric standpoint, my calculus professor is insistent that we use the more algebraic approach in our answers. However, I shall attempt to run this by her should I get the chance. $\endgroup$ – Overclock May 15 '16 at 17:31
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    $\begingroup$ @Overclock Look at it this way: the simple geometric approach gives you an easy way to check your work. $\endgroup$ – amd May 15 '16 at 18:27
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This is a bad exercise, since the answer is much easier to get by noting that we're looking for the points closest to and farthest from the origin on the sphere of radius $1$ around the point $u=(3,2,1)$, which are $u\pm\frac u{|u|}$.

If you really want to do it the hard way, you can solve, say, your third equation for $\lambda$, substitute that into the first and second equation and then substitute from that into the constraint, yielding a quadratic equation for $z$.

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