Question

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x1, x2, ..., xn) = x1 + x2 + ... + xn; x12 + x22 + ... + xn2 = 4

Answer 1

`f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_(i=1)^nx_i`

`{x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_(i=1)^n{x_i}^2=4`

The Lagrangian is

`L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_(i=1)^nx_i+\lambda\left(\sum_(i=1)^n{x_i}^2-4\right)`

with partial derivatives (all set equal to 0)

`L_(x_i)=1+2\lambda x_i=0\implies x_i=-\frac1{2\lambda}`

for
`1\le i\le n`, and

`L_\lambda=\displaystyle\sum_(i=1)^n{x_i}^2-4=0`

Substituting each
`x_i` into the second sum gives

`\displaystyle\sum_(i=1)^n\left(-\frac1{2\lambda}\right)^2=4\implies\frac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4`

Then we get two critical points,

`x_i=-\frac1{2\frac{\sqrt n}4}=-\frac2{\sqrt n}`

or

`x_i=-\frac1{2\left(-\frac{\sqrt n}4\right)}=\frac2{\sqrt n}`

At these points we get a value of
`f(x_1,\cdots,x_n)=\pm2\sqrt n`, i.e. a maximum value of
`2\sqrt n` and a minimum value of
`-2\sqrt n`.