Min/Max with Lagrange Multipliers

Question

Find the max and min values of the function $f(x,y,x)=3x+2y+4z$ with constraint $g(x,y,z) = x^2+2y^2+6z^2=16.$

I set $\nabla f = \lambda \nabla g$.

This gives me

$3=λ2x$,

$2=λ4y$,

$4=λ12z$ .

Now, setting $λ$s equal to each other I get that $x=3y$ and $z=(2/3)y.$ I plug these back into $g$, and solve for $y$.

However, I keep getting the wrong solution. I'm not sure where I'm going wrong...

Answer 1

If you know Cauchy-Schwarz inquality, it's a bit faster:

$\left|3x+2y+4z\right| =\left|3x + \sqrt{2}\cdot \left(\sqrt{2}y\right) + \dfrac{4}{\sqrt{6}}\cdot \left(\sqrt{6}z\right)\right| \leq \sqrt{9+2+\dfrac{8}{3}}\cdot \sqrt{x^2+ 2y^2+ 6z^2} = \dfrac{4\sqrt{123}}{3}$

Answer 2

Your method is correct. It is possible that the solution key is incorrect. You should be getting:

$ x = \pm 12 \sqrt{\frac{3}{41}} \\ y = \pm 4 \sqrt{\frac{3}{41}} \\ z = \pm \frac{8}{\sqrt{123}} $