Find the points on the surface y² = 64 + xz that are closest to the origin. 
(x, y, z) = ( ) (smaller y-value)
(x, y, z) = ( ) (larger y-value)

Solution:

The distance between an arbitrary point on the surface and the origin is

d(x, y, z) = √x² + y² + z²

Here we have to minimize x² + y² + z² to y² = 64 + xz, it's easy to show that √f(x) and f(x) share the same critical points.

We can consider augmented distance function, D(x,y,z) = √x² + y² + z²

Using Lagrange multipliers,

L(x, y, z, λ) = x² + y² + z² + λ(y² - 64 - xz)

We have partial derivatives

L_x = 2x - λz

L_y = 2y + 2yλ

L_z = 2z - λx

L_λ = y² - 64 - xz

Set each partial derivative to zero to find critical points.

Lᵧ = 0

⇒ 2y + 2yλ = 0

⇒ 2y( 1 + λ) = 0

y = 0 and λ = -1

Put λ = -1 in L_x and L_z

L_x = 0

⇒ 2x - λz = 0

⇒ 2x + z = 0 --- (1)

L_z = 0

⇒ 2z - λx = 0

⇒ 2z + x = 0 --- (2)

Solving (1) and (2)

We get, x = 0 and z = 0

Put the values of x and z in y² = 64 + xz

This means y² = 64

y = ±8

Therefore, the points on the surface closest to the origin are (0, ±8, 0).

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Find the points on the surface y² = 64 + xz that are closest to the origin. 

Summary:

The points on the surface y² = 64 + xz that are closest to the origin are (0, ±8, 0).