Max and Min Values - Lagrange Multipliers and 2 Constraints

Question

I'm asked to find the minimum and maximum values of $f(x, y, z) = x^2+y^2+z^2$ given the constraints $x+2y+z=5$ and $x-y=6$.

I have successfully computed the following: $x = \frac{57}{11}, y = \frac{-9}{11}, z= \frac{16}{11}$.

I was then able to obtain $f(\frac{57}{11}, \frac{-9}{11}, \frac{16}{11}) = \frac{326}{11}$.

This is my minimum value. However, I am unable to determine how to solve for a maximum value. Any ideas?

Answer 1

The two constraints yield the set of points on a line.

Since a line is unbounded, there is no maximum.

Answer 2

Rewrite the constraints and substitute \begin{eqnarray*} x&=& y+6 \\ z&=& -1-3y \\ f(x,y,z)&=& (y+6)^2+y^2+(-1-3y)^2=11y^2+18y+37= 11 \left(y+\frac{9}{11} \right)^2+\frac{326}{11}. \end{eqnarray*} So you have found the minimum ... now what can you say about the maximum ?

Answer 3

For this particular problem, it might be helpful to consider what is being actually asked. The question can be framed as follows:

Find the maximum and minimum radius of the sphere which contains the line of intersection of the planes $x+2y+z=5$ and $x-y=6$.

Now the two planes intersect at a line with the equation $3y+z+1=0$. The question asks us to find the minimum and maximum values of radii of the sphere that has a point on the given line.

It is obvious that there is no maximum value, as the line continues infinitely, whereas the minimum value is simply the shortest distance of the line from the origin.