What is the shortest distance from the surface 𝑥𝑦+6𝑥+𝑧^2=36 to the origin?

Answer 1

Let (x ,y, z) any point on surface.

We have to minimize function f(x,y,z) = x² + y² + z² under condition: xy + 6x + z² - 36 = 0;

Lagrange funct F(x,y,z,λ) = x² + y² + z² + λ(xy + 6x + z² - 36)

F'_x = 2x + λy + 6λ = 0;

F'_y = 2y + λx = 0

F'_z = 2z + 2λz = 0; z = 0 or λ = - 1;

F'_λ = xy + 6x + z² - 36 = 0

λ = -1; 2x - y - 6 =0 and 2y - x = 0 x = 2y; 4y - y - 6 = 0; y = 2, x = 4; 8 + 24 + z² - 36 = 0; z = ± 2

We get 2 points: (4, 2, 2) and (4, 2, - 2);

If z = 0, 2y² + 12y - 36 = 0; y² + 6y - 18 = 0; y = -3 ± 3√3 and x = - 6 ± 6√3

We get 2 more points: (- 6 + 6√3, - 3 + 3√3, 0) and (- 6 - 6√3 - 3 - 3√3, 0)

Calculating F(x,y,z) at all this points to get min. We have this 2 points: (4,2,2) and (4,2,-2).

f(4,2,2) = f(4,2, - 2) = 24 AND N MIN DISTANCE FROM ORIGIN IS √24