Find the extrema of f subject to the stated constraints. f ( x , y ) = x − y, subject to x 2 − y 2 = 2I'm solving a problem involving Lagrange's multipliers, and I've got this system of equations to solve: 1 = λ 2 x − 1 = − λ 2 y x 2 − y 2 − 2 = 0

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Find the extrema of   f subject to the stated constraints.  f  (  x  ,  y  )  =  x  −  y, subject to       x    2    −      y    2    =  2I&#039;m solving a problem involving Lagrange&#039;s multipliers, and I&#039;ve got this system of equations to solve:  1  =  λ  2  x  −  1  =  −  λ  2  y      x    2    −      y    2    −  2  =  0

aitantiskbx2v · Accepted Answer

You have  1  =  λ  2  xYou can multiply by   −  y so that the right side contains the right side from the second equation.  −  y  =  x  (  −  λ  2  y  )Then make use of the second equation  −  y  =  x  (  −  1  )So   y  =  x. Put that in y our third equation:      x    2    −      x    2    −  2  =  0and there is a problem. This says   −  2  =  0, so there are no solutions. In the context of your original problem, that makes sense. The curve you are restricted to is a hyperbola, and the line   y  =  x is one of its asymptotes. There will be no place where the function takes extremal values. Either you can keep moving along the hyperbola closer and closer to that asymptote, with   f increasing the whole time, or you can move away from the asymptote forever with   f always reducing.Another way to say that is, the gradient of   f is never parallel to this hyperbola.

Find the extrema of f subject to the stated constraints. f ( x , y ) = x &#x

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