Optimize f(x,y,z) = 2x + 2y + 2z

[squenshl]

Optimize f(x,y,z) = 2x + 2y + 2z subject to constraints g(x,y,z) = x² - y² - z² - 1 = 0 & h(x,y,z) = x² + y² + z² - 17 = 0.

I found L_x = 2 + 2\lambdax - 2\mux = 0, L_y = 2 + 2\lambday - 2\muy = 0 & L_z = 2 + 2\lambdaz - 2\muz = 0
I'm just asking how to use L_x, L_y & L_z to eliminate \lambda & \mu to get the critical points.

 

[HallsofIvy]

Your "Lx" is incorrect. Because x² is positive in both f and g, it should be 2+ 2\lambda x+ 2\mu x= 0

Those reduce very easily to (\mu+ \lambda)x= -1, (\mu- \lambda)y= -1 and (\mu- \lambda)z= -1. Also, just adding the two constraints eliminates both y and z giving 2x^2- 18= 0 so x= \pm 3. Divide the second equation by the third to get y/z= 1 so y= z. Put that into x^2- y^2- z^2- 1= 0 to get 9- 2z^2= 1 so 2z^2= 8 and z= \pm 2. The function is optimized at (3, 2, 2), (-3, 2, 2), (3, -2, -2), and (-3, -2, -2).

 

[squenshl]

But I thought L(x,y,z,\lambda,\mu) = 2x + 2y + 2z - \lambda(x² - y² - z² - 1) - \mu(x² + y² + z² - 17) making x² both negative making L_x = 2 - 2\lambdax - 2\mux. But I don't think it matters in the end result.

 

[squenshl]

What if I wanted to find \lambda & \mu?

 

[HallsofIvy]

But I thought L(x,y,z,\lambda,\mu) = 2x + 2y + 2z - \lambda(x² - y² - z² - 1) - \mu(x² + y² + z² - 17) making x² both negative making L_x = 2 - 2\lambdax - 2\mux. But I don't think it matters in the end result.

It just changes the values of \lambda and \mu.

What if I wanted to find \lambda & \mu?

Why would you? They are never part of the original problem. But if you do, once you have found x, y, and z, you can substitute those back into the equations to find \lambda and \mu.