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Use the method of Lagrange multipliers to optimize $f$ as indicated, subject to the given constraint(s).Find the critical points of $f(x, y)=x^{2}+24 x y+8 y^{2}$ such that $x^{2}+y^{2}=25$

$(-3,4,425),(3,4,425),(-4,3,-200),(4,-3,-200)$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 4

The Method of Lagrange Multipliers

Partial Derivatives

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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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now in this problem we have a just we don't have a scenario here. We just want to find the critical points of x squared plus 24 X Y plus eight Y squared subject to the constraints that X squared plus y squared minus 25 0. So this is a constraint that we're on a sphere of five. The radius of five. I'm sorry. A circle great is five as we're Justin to be here. So it's a former R G function. All right. With the Lagrange multiplayer and we can then start taking partial. So we only have three variables in this case, which is nice because it's kind of ugly. So we have this is first equation, second equation and then our third equation is here. And so you can see those are non linear and so we're gonna have multiple solutions. And if we go through and just keep very careful with the algebra, we'll find we'll find four solutions. We'll have um X equals -4. Y It was three and 8. X equals minus three, Y equals minus four is minus 17. Actually it was three. White was four. Land equals 17 and x equals four, Y equals minus three and lambda equals eight again. So these three these four solutions give values of -200 for these two solutions and 4 25 for these two solutions. So this would be our minimums and these are maximums. And what we can do is we can solve this and I solved it for let's see, solved it for why this is actually X. And this is f Bar. So I solved this for a while and substituted in here and then plotted that and again we have two branches here, so we have plus or minus here, so we have two different values of F. Bar so this is one of them and this is the other one. So we basically get this loop here and we can see here that we have a maximum Period about four, And when X is three and again X can be minus three. Also I should have actually called the negative values would have, you know, the rest of this, this is mirrored over here And then we can see here at four we have this value when we have a minimum at -200 and that would also be metered over here -4. So we can see our to maximum world, we can only see one of them, but this is mirrored about this axis, so that's what, you know, graphically geometrically, we can look at and make sure that again if we embed the constraint in our function and then maximize that new function or extra mayes it, we get the same solutions

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