Respuesta :
Answer:
Maximum value: [tex] 3* \sqrt{n} [/tex]
Minimum value: [tex] -3* \sqrt{n} [/tex]
Step-by-step explanation:
Let [tex] g(x) = x_1^2 + x_2^2+x_3^2+ ----+ x_n^2[/tex] , 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:
[tex] \nabla{f}(x_1, ..., x_n) = \lambda \nabla{g} (x_1,...,x_n) [/tex]
for a constant [tex] \lambda [/tex] .
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
[tex] 1 = \lambda 2 x_1 [/tex]
Thus,
[tex] x_i = \frac{1}{2\lambda} = c [/tex]
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
[tex] c = \, ^+_- \, \sqrt{\frac{9}{n} } = \, ^+_- \frac{3}{\sqrt{n}} [/tex]
When c is positive, f reaches a maximum, which is [tex] \frac{3}{\sqrt{n}} + \frac{3}{\sqrt{n}} + \frac{3}{\sqrt{n}} + ..... + \frac{3}{\sqrt{n}} = n * \frac{3}{\sqrt{n}} = 3 * \sqrt{n} [/tex]
On the other hand, when c is negative, f reaches a minimum, [tex]-3 * \sqrt{n} [/tex]
