Question

Find the extreme values of f subject to both constraints. (if an answer does not exist, enter dne.) f(x, y, z) = x + 2y; x + y + z = 4, y2 + z2 = 4

Answer 1

Using the method of Lagrange multipliers, we have the Lagrangian

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

with partial derivatives (set equal to 0)

$L_x=1+\lambda_1=0\implies\lambda_1=-1$
$L_y=2+\lambda_1+2\lambda_2y=0\implies\lambda_2y=-\dfrac12$
$L_z=\lambda_1+2\lambda_2z=0\implies\lambda_2z=\dfrac12$
$L_{\lambda_1}=x+y+z-4=0$
$L_{\lambda_2}=y^2+z^2-4=0$

From $L_y$ and $L_z$, we find that $\lambda_2y=-\lambda_2z\implies y=-z$. Then substituting into $L_{\lambda_2}$, we find

$y^2+z^2=4\implies2y^2=4\implies y=\pm\sqrt2\implies z=\mp\sqrt2$

and substituting these into $L_{\lambda_1}$, we get

$x+y+z=4\implies x=4$

So we have two possible critical points, $(4,\pm\sqrt2,\mp\sqrt2)$, which give extreme values of $4+2\sqrt2$ and $4-2\sqrt2$, respectively.