Find the points on the surface xy - xz^2 + 3yz = 0 that are closest to the point (1, -1, 1) by using Lagrange's multiplier method.

Steps to Solve

1. Define the objective function

To find the points on the surface closest to (1, -1, 1), we need to minimize the square of the distance from a point $(x, y, z)$ on the surface to $(1, -1, 1)$.

The distance squared is given by:

$$ D = (x - 1)^2 + (y + 1)^2 + (z - 1)^2 $$

2. Define the constraint

The given surface equation is:

$$ g(x, y, z) = xy - xz^2 + 3yz = 0 $$

This will be our constraint in the optimization problem.

3. Set up Lagrange's multiplier

We need to set up the Lagrange function:

$$ \mathcal{L}(x, y, z, \lambda) = D + \lambda g(x, y, z) $$

Substituting the expressions for $D$ and $g$, we get:

$$ \mathcal{L}(x, y, z, \lambda) = (x - 1)^2 + (y + 1)^2 + (z - 1)^2 + \lambda (xy - xz^2 + 3yz) $$

4. Calculate the gradients

We need to find the partial derivatives of $\mathcal{L}$ with respect to $x$, $y$, $z$, and $\lambda$, and set them to zero.

The partial derivatives are:

$$ \frac{\partial \mathcal{L}}{\partial x}, \frac{\partial \mathcal{L}}{\partial y}, \frac{\partial \mathcal{L}}{\partial z}, \frac{\partial \mathcal{L}}{\partial \lambda} $$

5. Set the partial derivatives to zero and solve

Set each of the partial derivatives to zero:

$$ \frac{\partial \mathcal{L}}{\partial x} = 2(x - 1) + \lambda (y - z^2) = 0 $$

$$ \frac{\partial \mathcal{L}}{\partial y} = 2(y + 1) + \lambda (x + 3z) = 0 $$

$$ \frac{\partial \mathcal{L}}{\partial z} = 2(z - 1) + \lambda (-2xz + 3y) = 0 $$

$$ g(x, y, z) = 0 $$

Solve these equations simultaneously to find $x$, $y$, $z$, and $\lambda$.

6. Solve the system of equations

This step involves substituting values from one equation into another and solving for the three variables, typically leading to a system of algebraic equations.