Each of these extreme value problems 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(x,y)=X^2-y^2;x^2+y^2=1

College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter5: Systems Of Equations And Inequalities
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Each of these extreme value problems 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(x,y)=X^2-y^2;x^2+y^2=1

Expert Solution
Step 1

Lagrange's method of multipliers

For a function of two or more variables this method is used.

Let f(x, y)=0 be the function within the condition h(x, y)=0.

First evaluate partial derivative of function and the constraint with respect to x and y and then obtain Lagrange's equation

fx+khx=0fy+khy=0

Evaluate x and y in terms of k and substitute these values in the given condition to obtain k and then x and y which do not have k in their value.

Put value of x and y in the function to obtain minimum or maximum.

 

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