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Use the method of Lagrange multipliers to optimize $f$ as indicated, subject to the given constraint(s).Minimize $f(x, y)=x y$ such that $y=\frac{4 x}{x-3},$ where $x>3$

48

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 4

The Method of Lagrange Multipliers

Partial Derivatives

Harvey Mudd College

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Lectures

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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01:28

Use the method of Lagrange…

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01:45

Use Lagrange multipliers t…

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01:44

mm Here we want to um uh We want to minimize uh huh. F equals X times Y. Subject to the constraint that y minus four, X over X minus three equals zero. So in this problem we have let's see here we have um Okay, are augmented equation are augmented function is X times Y plus land at times. See as always that's just how the lagrange multiplier method works. And now we take partial with respect to X gets a little ugly and we wound up with 12 lambda plus X minus three squared times Y. All over x minus three squared. And we set that equal to zero. And then with respect to Y one that one's easy. We get lambda plus X equals zero. And with respect to lambda we just get C equals zero. So we can um basically, you know lambda in terms of uh lambert in terms of X. Here we can get lambda in terms of X. And what we can plug that into here. Get Y. In terms of X. Plug that into here. And then solved. It turns out we only have one solution actually. And that solution is when x equals six And y equals eight Planned. Happens to be -6 with that solution and plugging that in. And the athletes of course 48. So that is our Minimum value is 48 of this function subject to this constraint here

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