# Lagrange Multiplier in 3D

Find the minimum and maximum values of the function $f(x,y,z) = x+2y+3z$ where $(x,y,z)$ is on the sphere $x^2+y^2+z^2=1$ using Lagrange multiplier.

So I put them into the Lagrange form and got $L(x,y,z,\lambda) = x+2y+3z+ \lambda(x^2+y^2+z^2-1)$ And you then get simultaneous equations $L_x = 1+ 2\lambda x$, $L_y = 2+ 2\lambda y$, $L_z = 3+ 2\lambda z$, $L_\lambda = x^2 +y^2 + z^2-1$.

However, when I solved these simultaneously, I got some very strange solutions, so I don't know if I've done something wrong up to this point on my solving was wrong.

Thanks

## 2 Answers

I do not see strange results. Using the derivatives with respect to $x,y,z$, you then have $$x=-\frac{1}{2 \lambda }\qquad y=-\frac{1}{\lambda }\qquad z=-\frac{3}{2 \lambda }$$ So $$L_\lambda=\frac{7}{2 \lambda ^2}-1=0$$ which makes $$\lambda=\pm \sqrt{\frac{7}{2}}$$ Replacing, you then have $$x=\pm \frac{1}{\sqrt{14}}\qquad y=\pm\frac{2}{\sqrt{14}}\qquad z=\pm\frac{3}{\sqrt{14}}$$ to which correspond values of $\pm \sqrt{14}$.

• It might be good to clarify that the signs of $x$, $y$ and $z$ must agree. – amd Apr 14 '16 at 17:18

$x+2y+3z\leq\sqrt {x^2+y^2+z^2}\sqrt {1+4+9} = \sqrt {14}$, by Cauchy-Schwarz, so $\max$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = \dfrac{1}{\sqrt{14}}$ and $\min$ occurs at $x = \dfrac{y}{2} = \dfrac{z}{3} = -\dfrac{1}{\sqrt{14}}$. You should get a same answer if you solve the Lagrange equations correctly.

• Except that 1+4+9=14 not 10! – almagest Apr 14 '16 at 12:12