Lagrange multipliers, help!!

Question

a) Write down the equations to find the point on the surface (x+y)z²=1 in the first octant, closest to the origin, using the method of Lagrange multipliers.

b) Solve these equations to find the closest points

Answer 1

let the function to be minimized = F= x²+y²+z² and the constraint to be (x+y)z²-1=G. Thus F_x=λGx or 2x=λz² and 2y=λz² and 2z=λ(2z)(x+y) and G for part A. Part B, it can be seen that x=y and z²= 1/(x+y) and λ=1/(x+y). Solving yields x=y=z=0

Answer 2

a) Let f(x,y,z) is square of distance, it will be the same minimum as for distance.

f(x,y,z) = x² + y² + z² min under condition (x + y)z² - 1 = 0 (x≥0, y≥0, z≥0)

Lagrange function:

F(x,y,z,µ) = x² + y² + z² - µ[(x + y)z²- 1].

F_x = 2x - µz² = 0; F_y = 2y - µz² = 0; F_x = 2z - 2µ(x + y)z = 0; F_µ = (x + y)z² - 1 = 0

x = y = µz²/2; z[1 - µ(x + y)] = 0; z = x = y = 0 not satisfy last equation.

So, 1 - µ(x + y) = 0. x + y = 1/µ; µz² = 1/µ; z² = 1/µ². 1/µ·1/µ² - 1 = 0; µ = 1; x = y = z²/2;

z⁴ = 1; z = 1; x = y = 1/2. So, point is (1/2, 1/ 2, 1). Distance = (1/4 + 1/4 + 1)^1/2 = (3/2)^1/2

To show that this is minimum we take different point: (1, 3, 1/2) on thissurface. Then distance

d = (1 + 9 + 1/4)^1/2 = (41/4)^1/2> (3/2)^1/2.