Question

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints. (If a value does not exist, enter NONE.) f(x,y,z) = 3x – y – 3z; x + y – z = 0, x2 + 2z2 = 1

Answer 1

The Lagrangian is

$L(x,y,z,\lambda,\mu)=3x-y-3z+\lambda(x+y-z)+\mu(x^2+2z^2-1)$

with critical points where its partial derivatives vanish:

$L_x=3+\lambda+2\mu x=0$

$L_y=-1+\lambda=0\implies\lambda=1$

$L_z=-3-\lambda+4\mu z=0$

$L_\lambda=x+y-z=0$

$L_\mu=x^2+2z^2-1=0\implies x^2+2z^2=1$

Since we know $\lambda=1$, we have

• $L_x=0\implies4+2\mu x=0\implies x=-\dfrac2\mu$
• $L_z=0\implies-4+4\mu z=0\implies z=\dfrac1\mu$
• $L_\mu=0\implies\dfrac6{\mu^2}=1\implies\mu=\pm\sqrt6$, so that $x=\mp\dfrac2{\sqrt6}$ and $z=\pm\dfrac1{\sqrt6}$
• $L_\lambda=0\implies y=z-x\implies y=\pm\dfrac3{\sqrt6}$

So there are two critical points, $\left(-\dfrac2{\sqrt6},\dfrac3{\sqrt6},\dfrac1{\sqrt6}\right)$ and $\left(\dfrac2{\sqrt6},-\dfrac3{\sqrt6},-\dfrac1{\sqrt6}\right)$, which give a minimum value of $-2\sqrt6$ and a maximum value of $2\sqrt6$, respectively.