Use lagrange multiplier techniques to find the local extreme values of f(x, y) = x2 − y2 − 2 subject to the constraint x2 + y2 = 16

Respuesta :

Given [tex]f(x,\ y)=x^2-y^2-2[/tex] subject to the constraint [tex]x^2+y^2=16[/tex]

Let [tex]g(x,\ y)=x^2+y^2[/tex].

The gradient vectors of [tex]f[/tex] and [tex]g[/tex] are:

[tex]\nabla f(x,\ y)=\langle2x,-2y\rangle[/tex] and [tex]\nabla g(x,\ y)=\langle2x,2y\rangle[/tex]

By Lagrange's theorem, there is a number [tex]\lambda[/tex], such that

[tex]\langle2x,-2y\rangle=\lambda\langle2x,2y\rangle=\langle2\lambda x,2\lambda y\rangle[/tex]

[tex]\lambda=\pm1[/tex]

It can be seen that [tex]f(x,\ y)=x^2-y^2-2[/tex] has local extreme values at the given region.