Determine local extreme points, minimum and maximum

Question

I've been working on this problem for hours and just can't solve it.

$f : R^3 \to R$, $f(x) = 4x_1 + x_2 - 2x_3$ on the quantity $K = \{x\in R^3: x_1 + x_2 + x_3 = 0, x_1^2 + x_2^2 + x_3^2 = 1\}$.

I was told to do an equivalent transformation, so I divided by f on both sides and got $\{4x_1 + x_2 - 2x_3 / f\}$, but I don't know how that will help me solve the problem.

Answer 1

Maybe you could try something like $$f(x) = 4x_1+x_2-2x_3 = 4x_1+x_2 + x_3 - 3x_3 = 3x_1-3x_3 = 3(x_1-x_3)$$ for all $x \in K$. This reduces the problem to finding the min/max of $x_1-x_3$ on $K$. I think the min on $K$ is $(-\frac{\sqrt{2}}{2},0,\frac{\sqrt{2}}{2})$ and the max on $K$ is $(\frac{\sqrt{2}}{2},0,-\frac{\sqrt{2}}{2})$.

EDIT: I am not sure if the min and max I listed are correct.

Answer 2

Probably not what was intended:

Note that $(1,0,-1), (1,-2,1)$ span $\{ x | \sum_k x_k = 0 \}$. Let $v_1 = { 1\over \sqrt{2}} (1,0,-1), v_2 = {1 \over \sqrt{6}} (1,-2,1)$, note that $v_1 \bot v_2$.

Then with $x= (\cos t) v_1 + (\sin t) v_2$ we see the constraints are implicitly satisfied and we can look at $(\cos t) f(v_1) + (\sin t) f(v_2)$, evaluating gives $3 \sqrt{2} \cos t$. Hence the $\max,\min$ occur at $\pm v_1$.

Answer 3

This is standard example of Lagrange's method of multipliers.

Taking $g = 4x_1+x_2-2x_3+\lambda_1(x_1+x_2+x_3)+\lambda_2(x_1^2+x_2^2+x^3-1)$

Stationary points are given by $\frac{\partial g}{\partial x_1}=0, \frac{\partial g}{\partial x_2}=0, \frac{\partial g}{\partial x_3}=0,\frac{\partial g}{\partial \lambda_1}=0, \frac{\partial g}{\partial \lambda_2}=0$

From first 3 equations, we get $$x_1=-\frac{4+\lambda_1}{2\lambda_2}, x_2=-\frac{1+\lambda_1}{2\lambda_2}, x_1=-\frac{-2+\lambda_1}{2\lambda_2}$$

These can be substituted in last 2 equations to get $\lambda_1=-1, \lambda_2=\pm\frac{3}{\sqrt 2}$ (if my calcs are right).

From here on, you can find $x_1, x_2, x_3$ and the corresponding maxima/minima.

I am getting maxima $f=3\sqrt 2$ at $(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}})$, minima $f=-3\sqrt 2$ at $(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}})$