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Use the method of Lagrange multipliers to optimize $f$ as indicated, subject to the given constraint(s).Maximize $f(x, y, z)=x y z$ such that $3 x+2 y+5 z=27$

$243 / 10$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 4

The Method of Lagrange Multipliers

Partial Derivatives

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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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Use the method of Lagrange…

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07:24

Use Lagrange multipliers t…

01:59

mhm. This is very similar to the previous problem where we have the function that we want to let's see here maximize, I probably think maximize this X times Y times Z subject to this constraint, which is a plane. So we take this function and slice it the plane. And we try to find the maximum on that plane. Now are augmented. Lagrange are augmented function for the uh Lagrange multiplier method, um is X times Y times Z plus lambda times this constraint here. And then we can take partials with respect to X, Y. Z. And lambda with respect to X. We get three lambda plus Y. Z. And that equals zero, expect to why we get to lander plus X. E. And we set that to zero, expect to see we have five lambda plus X times Y. Set that equal to zero. And then we expect the lamb that we just get our constraint function back and we set that equal to zero to get our constraint equation. And now again we have the same kind of things as we did in the last problem, there's three trivial solutions where two of x, Y and z are zero, so, and then the other satisfies this. So we get 00 X zero, Y zero and Z is 27/5, okay, X zero, Y is 27/2, 00 and X is nine, Y is 000. λ is zero in all those cases. Now we plug those back into into here and we get zero for all those, right? Because if X, Y or Z is zero, then we have zero for the function. Then we find 11 like nontrivial solution, and that is excess three, Y is nine over to And Z is 9/5 and 50. And we take this and we plug it into here and we get to 43/10. And so that is going to be a R maximum.

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