Determine the Conditional Extremes of a Function

Question

I am trying to determine the conditional extremes the following question:

• Determine the point of the plane, $2x-y+2z=16$ closest to the origin.

but I do not fully understand the question. If I am interpreting the question correctly then:

1. The function would be: $f(x,y,z) = 2x+2y+2z=16$.
2. The condition would be something like $x + y = ?$.

But I am not sure how exactly to interpret the condition from the question. Any suggestions?

Answer 1

The distance of a point $(x,y,z)$ in the plane and the origin, $(0,0,0)$, is equal to

$$ ||(x,y,z)|| = \sqrt{x^2+y^2+z^2} .$$

So the problem you describe can be written as

$$ \min_{x,y,z} x^2+y^2+z^2\quad\text{subject to} \quad 2x−y+2z=16.$$

The first order conditions are: \begin{align} [x]:&\quad 2x+2\lambda=0\\ [y]:&\quad 2y-\lambda=0\\ [z]:&\quad 2z+2\lambda=0 \end{align} where $\lambda$ is the Lagrange multiplier. Solving this system yields $$ x=-\lambda,\quad y=\frac{\lambda}{2},\quad\text{and}\quad z=-\lambda.$$

Substituting this into the constraint we get

$$-2\lambda-\frac{\lambda}{2}-2\lambda=16 \Rightarrow \lambda=-\frac{32}{9}.$$

Finally, the point can be solve for by plugging $\lambda$ back: $$ x=\frac{32}{9},\quad y=-\frac{32}{18},\quad\text{and}\quad z=\frac{32}{9},$$

and the associated distance is $$\sqrt{x^2+y^2+z^2}=\frac{16}{3}. $$

Answer 2

the Hessian Normalform of your equation is given by $$\frac{2x-y+2z-16}{\pm3}$$ Setting $P(0;0;0)$ in the given term we get the disatnce to the origo to the plane this is given by $$\frac{16}{3}$$