Question

We need to optimize f(x, y, z) = x + 2y subject to the constraints g(x, y, z) = x + y + z = 8 and h(x, y, z) = y2 + z2 = 4. To find the possible extreme value points, we must use ∇f = λ∇g + μ∇h.

Answer 1

The Lagrangian is

$L(x,y,z,\lambda,\mu)=x+2y+\lambda(x+y+z-8)+\mu(y^2+z^2-4)$

with critical points where the partial derivatives vanish:

$L_x=1+\lambda=0\implies\lambda=-1$

$L_y=2+\lambda+2\mu y=0\implies\mu=-\dfrac1{2y}$

$L_z=\lambda+2\mu z=0\implies\mu=\dfrac1{2z}$

$L_\lambda=x+y+z-8=0$

$L_\mu=y^2+z^2-4=0$

The second and third equations tell us $z=-y$, so that in the last equation we find

$y^2+(-y)^2=2y^2=4\implies y=\pm\sqrt2\implies z=\mp\sqrt2$

and from the fourth equation we get

$x+y+z=x=8$

So we have two critical points, $(8,\sqrt2,-\sqrt2)$ and $(8,-\sqrt2,\sqrt2)$, which give respective extreme values of $f(8,\sqrt2,-\sqrt2)=8+2\sqrt2$ (maximum) and $f(8,-\sqrt2,\sqrt2)=8-2\sqrt2$ (minimum).