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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 $f(x, y, z)=x+y+z$ such that $x^{2}+y^{2}+z^{2}=36$

$(-2 \sqrt{3},-2 \sqrt{3},-2 \sqrt{3},-6 \sqrt{3}),(2 \sqrt{3}, 2 \sqrt{3}, 2 \sqrt{3}, 6 \sqrt{3})$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 4

The Method of Lagrange Multipliers

Partial Derivatives

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Harvey Mudd College

Boston College

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.

04:14

02:28

Use the method of Lagrange…

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03:12

02:48

03:53

02:20

02:06

03:05

the problem is very, very similar to the previous problem but now we're finding my critical points of X plus Y plus C. And now our constraint is a sphere of radius six. So we form our augmented function with the Lagrange multiplier. Take partials on satyam equals zero. So we get one plus two lambda X equals 01 plus two lambda Y equals 01 plus two lambda, Z equals zero and then C equals zero. So she is here in that equals zero. And we can then uh we then you know just back substitute and you know eliminate variables. And we finally find out that we had to solutions one is X. Y and Z are all two times the square to three. And lambda winds up being minus 1/4 times the square three. The other solution is X, Y and Z are all minus two times the square to three. And Lander winds up being 1/4 times the square three. So this plug that into here and we obviously get six times square to three. This gives us -6 times the square to three. So this is a maximum, this is a minimum. And if we if we um What I did is I solved this is for Z got you know the 22 solutions e equals group Z equals plus or minus something, plug that into here. So I get two branches of F and I plotted those two branches and again we get this kind of like Mylar but inflated balloon like structure here and it shows that again we can see where are, you know, we have a maximum over here and the minimum, basically it's underneath there. And those are, you know, we can see that they're on this kind of diagonal, um, that sequence by cuisine.

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