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bogdanovich [222]
3 years ago
11

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

treme values of the function subject to the given constraint. f(x1, x2, ..., xn) = x1 + x2 + ... + xn; x12 + x22 + ... + xn2 = 4
Mathematics
1 answer:
olga nikolaevna [1]3 years ago
6 0

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}

or

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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