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\\eq1` 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\frac1{√(1+4\mu^2)}`

`L_z=0\implies z=\mp\frac1{2\mu√(1+4\mu^2)}`

`L_\mu=0\implies\frac1{1+4\mu^2}+\frac1{4\mu^2(1+4\mu^2)}=1\implies\mu=\pm\frac12`

`L_\lambda=0\implies x=\pm√(1+4\mu^2)`

Now,

• if
  `\mu=\frac12`, we have two critical points at
  `\left(\sqrt2,\frac1{\sqrt2},-\frac1{\sqrt2}\right)` and
  `\left(-\sqrt2,-\frac1{\sqrt2},\frac1{\sqrt2}\right)`;
• if
  `\mu=-\frac12`, we have two additional critical points
  `\left(\sqrt2,\frac1{\sqrt2},\frac1{\sqrt2}\right)` and
  `\left(-\sqrt2,-\frac1{\sqrt2},-\frac1{\sqrt2}\right)`

The first two critical points give a minimum value of
`f(x,y,z)=\frac12`, and the other two give a maximum value of
`f(x,y,z)=\frac32`.