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

Maximum value:
`3* √(n)`

Minimum value:
`-3* √(n)`

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:

`\\abla{f}(x_1, ..., x_n) = \lambda \\abla{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 = (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{(9)/(n) } = \, ^+_- (3)/(√(n))`

When c is positive, f reaches a maximum, which is
`(3)/(√(n))  +  (3)/(√(n)) +  (3)/(√(n))  + ..... +  (3)/(√(n))  = n *  (3)/(√(n))  = 3 * √(n)`

On the other hand, when c is negative, f reaches a minimum,
`-3 * √(n)`