Answer to Question #211332 in Calculus for rahul

Question

use Lagrange’s Multiplier method to find the maximum and minimum of f(x,y)= y^2 - x^2 subjected to the constraint of 1/4 x^2 + y^2 = 1

Answer 1

$f_x=\lambda g_x,f_y=\lambda g_y$

$g(x,y)=x^2/4+y^2=1$

$-2x=\lambda x/2$

$2y=2\lambda y$

So, we have:

$f(0,1)=1,f(0,-1)=1$ - maximum value

$f(2,0)=-4,f(-2,0)=-4$ - minimum value