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Use the method of Lagrange multipliers to optimize $f$ as indicated, subject to the given constraint(s).Maximize $f(x, y)=x y$ such that $x^{2}+y^{2}=16$

8

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 4

The Method of Lagrange Multipliers

Partial Derivatives

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12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

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Use the method of Lagrange…

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Use Lagrange multipliers t…

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we have F equals X times Y subject for the constraints that x squared plus y squared equals 16 or x squared plus y squared minus 16 is zero. So this is uh this means that x and y lie on a circle of radius four. Now our augmented equation is G equals X times Y plus lambert times X squared plus y squared minus 16. Yeah, take partial winters back to X. We get to lander X plus Y. Set that equal to zero. Take partial with respect to why we can X plus two lambda. Why set that equal to zero? Take partial with respect to the lambda. And we get C And that's equal zero. So now we have um you know, we can basically, so you know eliminate um say why and lambda from these uh get Y and lambda in terms of X from these two and then plug that into here and then get an equation for X. And what we see is we get we get to values um two solutions for X plus or minus square, two times the square to to and then we wind up with two values for each value of x. We get to values of why so plus or minus square to times square to to So in the end we wind up with four solutions and if we plug those in, you see that the solutions where we have the same signs in X and y are eight And we have different signs there -8 obviously. And so we wanted to maximize that. So we have to, we have to maximum and we have this is a maximum and this is a maximum and the and these are minimum.

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