Question

Find the minimum and maximum of f(x, y, z) = y + 4z subject to two constraints, 2x + z = 4 and x2 + y2 = 1. g

Answer 1

`L(x,y,z,\lambda_1,\lambda_2)=y+4z+\lambda_1(2x+z-4)+\lambda_2(x^2+y^2-1)`


`L_x=2\lambda_1+2\lambda_2 x=0\implies\lambda_1+\lambda_2x=0`

`L_y=1+2\lambda_2y=0`

`L_z=4+\lambda_1=0\implies\lambda_1=-4`

`L_(\lambda_1)=2x+z-4=0`

`L_(\lambda_2)=x^2+y^2-1=0`


`\lambda_1=-4\implies \lambda_2x=4\implies\lambda_2=\frac4x`

`1+2\lambda_2y=0\implies\lambda_2y=-\frac12\implies8y=-x`


`x^2+y^2=1\implies (-8y)^2+y^2=65y^2=1\implies y=\pm\frac1{√(65)}`

`y=\pm\frac1{√(65)}\implies x=\mp\frac8{√(65)}`

`2x+z=4\implies z=4\pm(16)/(√(65))`

We have two critical points to consider:
`\left(-\frac8{√(65)},\frac1{√(65)},4+(16)/(√(65))\right)` and
`\left(\frac8{√(65)},-\frac1{√(65)},4-(16)/(√(65))\right)`.

At these points, we respectively have a maximum of
`16+√(65)` and a minimum of
`16-√(65)`.