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Solve Example 4 by the alternate method suggested and verify that the critical point is indeed a minimum.

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 5

Economic Applications

Partial Derivatives

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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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Find the $x$ -coordinates …

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Find the critical points a…

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Find and classify the crit…

uh if you look closely at this problem uh maybe a pre calc teacher would explain this uh using like I guess telling you to factor things out. So if you were just looking at this function, I think you would know or agree with the possibility coefficient even agree. It's going to go up into the right uh and up until left. Now if you did this kind of a pre cal quay and you factored out and execute and you'll be left the next minus four. Uh what you would see then is that you have a solution of X equals zero, But with a multiplicity of three, it goes through the origin. So this x equals zero is not going to be a min or max, but exit goes for Well, right here would be a value that would be amended. Um So as we take this derivative, whether before x cubed -12 x squared And you set that equal to 0? Well, if you did the problem this way, you can factor out a four eps left with x squared minus nope. For x square, that would be the correct factor. And you'll be left with X -3 equals zero. So you're critical numbers. This is a X equals zero twice. Uh Multiple city of two. But then this one will be X equals three. So I would use the assigned chart of 003. And think of what happens with your zeros that if you were to plug in a number to the left of zero, you would get negatives for all of them. If you plug in, lets say one into these factors, you would get positives to the right of zero and negative to the left of three and then any value bigger than three, like four, you're going to get positives for all of these factors. So the negative times a negative times a negative is negative. So therefore when the first derivative is negative, what's decreasing positive, positive times negative is negative again decreasing. So X equals zero is not a min or max um but a positive times a positive times a positive positive. So we are increasing. The only way to go from decreasing to increasing Is that X equals three is a relative mint. X equals zero is not a man or a max.

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