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oksian1 [2.3K]
3 years ago
9

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

oes not exist, enter DNE.)
f(x1, x2, ..., xn) = x1 + x2 + ... + xn; x12 + x22 + ... + xn2 = 36
Mathematics
1 answer:
eduard3 years ago
4 0

The Lagrangian is

L(x_1,\ldots,x_n,\lambda_1,\ldots,\lambda_n)=x_1+\cdots+x_n+\lambda_1({x_1}^2+\cdots+{x_n}^2)+\cdots+\lambda_n({x_1}^2+\cdots+{x_n}^2)

with partial derivatives (set equal to 0)

\dfrac{\partial L}{\partial x_i}=1+2x_i(\lambda_1+\cdots+\lambda_n)=0

\dfrac{\partial L}{\partial\lambda_i}={x_1}^2+\cdots+{x_n}^2-36=0

for each 1\le i\le n.

Let \Lambda be the sum of all the multipliers \lambda_i,

\Lambda=\displaystyle\sum_{k=1}^n\lambda_k=\lambda_1+\cdots+\lambda_n

We notice that

x_i\dfrac{\partial L}{\partial x_i}=x_i+2{x_i}^2\Lambda=0

so that

\displaystyle\sum_{i=1}^nx_i\dfrac{\partial L}{\partial x_i}=\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0

We know that \sum\limits_{i=1}^n{x_i}^2=36, so

\displaystyle\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0\implies\sum_{i=1}^nx_i=-72\Lambda

Solving the first n equations for x_i gives

1+2\Lambda x_i=0\implies x_i=-\dfrac1{2\Lambda}

and in particular

\displaystyle\sum_{i=1}^nx_i=-\dfrac n{2\Lambda}

It follows that

-\dfrac n{2\Lambda}+72\Lambda=0\implies\Lambda^2=\dfrac n{144}\implies\Lambda=\pm\dfrac{\sqrt n}{12}

which gives us

x_i=-\dfrac1{2\left(\pm\frac{\sqrt n}{12}\right)}=\pm\dfrac6{\sqrt n}

That is, we've found two critical points,

\pm\left(\dfrac6{\sqrt n},\ldots,\dfrac6{\sqrt n}\right)

At the critical point with positive signs, f(x_1,\ldots,x_n) attains a maximum value of

\displaystyle\sum_{i=1}^nx_i=\dfrac{6n}{\sqrt n}=6\sqrt n

and at the other, a minimum value of

\displaystyle\sum_{i=1}^nx_i=-\dfrac{6n}{\sqrt n}=-6\sqrt n

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