Mathematics: 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

This extreme value problem has a solution with both

smartowl101

09.03.2022

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

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

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)

$L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}$

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

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

Then we get two critical points,

$x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}$

$x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}$

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 $-2\sqrt n$ .

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morris9878

01.02.2023

Step-by-step explanation:

The mean of a group of numbers is determined by adding all numbers together and then divided the sum by the total number of numbers.

For example: the mean of 1, 19, 100 = (1+19+100)/3 = 120/3 = 40

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