# Lagrange Multipliers with One Constraint Examples 6

Recall from The Method of Lagrange Multipliers page that with the Method of Lagrange Multipliers that if we have a function $z = f(x, y)$ subject to the constraint $g(x, y) = C$, then if extreme values exist and $\nabla g (x, y) \neq (0, 0)$ on the level curve $g(x, y) = C$ then we can obtain the critical points subject to the constraint by solving the system of equations below and then comparing the values of $f$ at these critical points.

(1)Furthermore, if we have a function $w = f(x, y, z)$ subject to the constraint $g(x, y, z) = C$, then if extreme values exist and $\nabla g(x, y, z) \neq (0,0,0)$ on the level surface $g(x, y, z) = C$, then we can find these extreme values by solving the system of equations below and then comparing the values of $f$ at these critical points.

(2)We will now look at some more examples of solving problems regarding Lagrange multipliers.

## Example 1

**Find the extreme values of the function $f(x_1, x_2, ..., x_n) = x_1 + 2x_2 + ... + nx_n$ subject to the constraint $x_1^2 + x_2^2 + ... + x_n^2 = 1$.**

Let $g(x_1, x_2, ..., x_n) = x_1^2 + x_2^2 + ... + x_n^2$. In calculating the necessary partial derivatives we have that:

(3)Note that $x_1, x_2, ..., x_n \neq 0$ otherwise the first $n$ equations would be contradicted. Thus we can divide freely by these variables to have that:

(4)This implies that $2x_1 = \frac{2x_n}{2} = ... = \frac{2x_n}{n}$. More generally, for each $j = 2, 3, ..., n$ we have that $x_j =jx_1 (*)$ and so plugging this information into the last equation and we have that

(5)Thus from $(*)$ we see that our points of interest are $\left ( \sqrt{ \frac{6}{n(n+1)(2n+1)}}, 2 \sqrt{ \frac{6}{n(n+1)(2n+1)}}, ..., n\sqrt{ \frac{6}{n(n+1)(2n+1)}} \right )$ and $\left ( - \sqrt{ \frac{6}{n(n+1)(2n+1)}}, -2\sqrt{ \frac{6}{n(n+1)(2n+1)}}, ..., -n\sqrt{ \frac{6}{n(n+1)(2n+1)}} \right )$.

Plugging these points into $f$ and we have that the maximum is achieved at the first point and is:

(6)Similarly, $f$ achieves a minimum at the second point and it is:

(7)