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

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

$L_x=2\lambda_1+2\lambda_2 x=0\implies\lambda_1+\lambda_2x=0$
$L_y=1+2\lambda_2y=0$
$L_z=4+\lambda_1=0\implies\lambda_1=-4$
$L_{\lambda_1}=2x+z-4=0$
$L_{\lambda_2}=x^2+y^2-1=0$

$\lambda_1=-4\implies \lambda_2x=4\implies\lambda_2=\dfrac4x$
$1+2\lambda_2y=0\implies\lambda_2y=-\dfrac12\implies8y=-x$

$x^2+y^2=1\implies (-8y)^2+y^2=65y^2=1\implies y=\pm\dfrac1{\sqrt{65}}$
$y=\pm\dfrac1{\sqrt{65}}\implies x=\mp\dfrac8{\sqrt{65}}$
$2x+z=4\implies z=4\pm\dfrac{16}{\sqrt{65}}$

We have two critical points to consider: $\left(-\dfrac8{\sqrt{65}},\dfrac1{\sqrt{65}},4+\dfrac{16}{\sqrt{65}}\right)$ and $\left(\dfrac8{\sqrt{65}},-\dfrac1{\sqrt{65}},4-\dfrac{16}{\sqrt{65}}\right)$.

At these points, we respectively have a maximum of $16+\sqrt{65}$ and a minimum of $16-\sqrt{65}$.