Question

Use the method of Lagrange Multipliers to End the maximum and minimum values of the function subject to the given constraints: f(x, y, z) = yz + xy xy = 1 y^2 + z^2 = 1

Answer 1

Looks like you're looking for the extrema of $f(x,y,z)=yz+xy$ subject to $xy=1$ and $y^2+z^2=1$. The Lagrangian is

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

Look for any critical points:

$L_x=y+\lambda y=y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_y=z+x+\lambda x+2\mu y=0$

$L_z=y+2\mu z=0$

$L_\lambda=xy-1=0\implies xy=1$

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

• If $y=0$, then

$L_z=0\implies2\mu z=0\implies z=0$ (we don't want $\lambda=\mu=0$)

but then $y^2+z^2=0\neq1$ so we omit this case.

• If $\lambda=-1$, then

$L_y=0\implies z+2\mu y=0$

$L_\mu=0\implies(1+4\mu^2)y^2-1=0\implies y=\pm\dfrac1{\sqrt{1+4\mu^2}}$

$L_z=0\implies z=\mp\dfrac1{2\mu\sqrt{1+4\mu^2}}$

$L_\mu=0\implies\dfrac1{1+4\mu^2}+\dfrac1{4\mu^2(1+4\mu^2)}=1\implies\mu=\pm\dfrac12$

$L_\lambda=0\implies x=\pm\sqrt{1+4\mu^2}$

Now,

• if $\mu=\dfrac12$, we have two critical points at $\left(\sqrt2,\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)$ and $\left(-\sqrt2,-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$;
• if $\mu=-\dfrac12$, we have two additional critical points $\left(\sqrt2,\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$ and $\left(-\sqrt2,-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)$

The first two critical points give a minimum value of $f(x,y,z)=\dfrac12$, and the other two give a maximum value of $f(x,y,z)=\dfrac32$.