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 = 9

Answer 1

Answer:

Maximum value: $3* \sqrt{n}$

Minimum value: $-3* \sqrt{n}$

Step-by-step explanation:

Let $g(x) = x_1^2 + x_2^2+x_3^2+ ----+ x_n^2$ , the restriction function.The Lagrange Multiplier problem states that an extreme (x1, ..., xn) of f with the constraint g(x) = 9 has to follow the following rule:

$\nabla{f}(x_1, ..., x_n) = \lambda \nabla{g} (x_1,...,x_n)$

for a constant $\lambda$ .

Note that the partial derivate of f respect to any variable is 1, and the partial derivate of g respect xi is 2xi, this means that

$1 = \lambda 2 x_1$

Thus,

$x_i = \frac{1}{2\lambda} = c$

Where c is a constant that doesnt depend on i. In other words, there exists c such that (x1, x2, ..., xn) = (c,c, ..., c). Now, since g(x1, ..., xn) = 9, we have that n * c² = 9, or

$c = \, ^+_- \, \sqrt{\frac{9}{n} } = \, ^+_- \frac{3}{\sqrt{n}}$

When c is positive, f reaches a maximum, which is $\frac{3}{\sqrt{n}} + \frac{3}{\sqrt{n}} + \frac{3}{\sqrt{n}} + ..... + \frac{3}{\sqrt{n}} = n * \frac{3}{\sqrt{n}} = 3 * \sqrt{n}$

On the other hand, when c is negative, f reaches a minimum, $-3 * \sqrt{n}$