Question

Find the extreme values of z on the surface 2x^2 + 3y^2 + z^2 – 12xy + 4xz = 35.(DO NOT USE LAGRANGE MULTIPLIERS)

Answer 1

A general concept: if f(x,y,z(x,y))=0 then differentiating with respect to x and y gets:

$f_x+f_zz_x = 0$

$f_y+f_zz_y = 0$

when $z_x = z_y = 0$ , you'll get $f_x = f_y = 0.$

So solve for the partials of f with respect to x and y are 0.

$f_x = 4x-12y+4z = 0$

$f_y = 6y-12x = 0$

solving for x,y as functions of z:

$x=z/5,y=2z/5$

Plug those into the original:

$2z^2/25+12z^2/25+z^2-24z^2/25+8z^2/25=35$

$23z^2=875$

$z=\pm 6.17$