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For $f(x, y, z)=2 x^{2}-16 x+3 y^{2}-24 y+4 z^{2}-40 z+25,$ find the point(s) at which $f_{x}(x, y, z)=0, f_{y}(x, y, z)=0,$ and $f_{z}(x, y, z)=0$.

$(4,4,5)$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

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Lectures

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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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So if we want to figure out where all of the partial derivatives of this R equals zero will first have, to find all those partials instead equals zero. So let's just go ahead and do that to start. So first I'll do del by Dell X to get the partial with respect to X. So now remember thes wise and these disease here and along with just, you know, this constant, we assume, are all just constant. So if I were to like, square it or raise a constitutive power and then multiply it, that's just going toe still be a concept. So if I take the derivative of all those, that's just gonna be zeros, all these end up just being zero and then we can go ahead and just use power will take the derivative here. So would be four x minus 16. All right, so now we want to set this equal to zero, and then we would go ahead and solve. So add 16. Divide by four. So that's X is equal to four. So what we have found is that the partial derivative with respect to X if we plug in four and then notice There's no restrictions on wires E. So it doesn't matter what we plug in for y Z, this will always be zero. Okay, uh, now we can go ahead and repeat this, but with the partial derivative for why And so now these exes and these disease along with that 25 are all constants. So we go ahead and just zero these out because again, the derivative of these since we're assuming there are constants, should just be zero. And so then that will give us the partial. But respect why is going to be equal to what we use power rule here? So it would be six. Why minus 24 and then we set this equal to zero. And so if we go ahead and salt for why that gives us, why is he before? Which implies that the partial respect why and so knows There's no restrictions on extra Z here, So it be x four z. So that would just be zero. And then, lastly, to get our partial with respect to see, we will go ahead, take the derivative with respect to see so we end up with F Z is equal to now again, the exes and the wise, we assume, are constants. So when we take the derivatives and all those just like in the previous two derivatives, those were just going to zero out and then we just use powerful for those. So that would be eight x minus 40. And then again, we said the secret to zero, which is going to imply. So we add 40 divide by eight. So Z is equal to five, which further implies that are partial derivative of prospective Z will be equal to zero as long as these five, and it doesn't matter what extra yr. All right, so now we actually go ahead and group all these together. Since we're interested in where this holds for all of those values, we would just take the intersection of all these solutions which would essentially just be like, Well, it says X has to be four. Why has to be four z has to be five. So then this implies for all of these to be zero at the same time, that's only going to happen at the 0.445 So that will be our point where all of our partials are zero

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