WEBVTT
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we're gonna need, uh, some graphing software that's
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able to do, um, slope field grafts.
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That's not on. Not all of them can,
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but one, but the one I'm using can.
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So I'm using Geo Deborah, if you need one
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. Um, and it's their c A s calculator
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. And if you'll notice on the left here,
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I've already plugged in a command that gives me the
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slow field for our system with C at 20.
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And I want you to look around at the plot
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and see if you can tell. Is there an
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equilibrium solution here? So scroll in and out a
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little bit, and I would say it looks like
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there is one appearing. It's right around 500.
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You see, there's limited resolution on these things,
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so sometimes you have to move around to spot it
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. But right around 500 you're seeing a pretty flat
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line. Okay, so it looks like minus 20
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Gifts Assn. Equilibrium solution. How about minus 10
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? Okay, so now it should be minus 10
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. So you're seeing there a couple interesting things here
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. First of all, you can probably eyeball it
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. It looks like there's an equilibrium solution. If
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I can just settle it on it. Sorry.
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I'm trying to get the well, Okay, maybe
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actually don't have to. So there's an equilibrium solution
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a little below 1000 and, um, right around
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500. No below 500. Maybe about 200.
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I'm estimating a bit there. Okay. And the
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way you can tell, even though there are no
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flatlines, is if you look a little bit of
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1000. There's this part where, um a bunch
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of all those slopes are pointing down and then immediately
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below them. They're pointing up. OK, so
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we don't from negative to positive. That means we
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crossed zero. So a little below 1000. There
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must be an equilibrium solution somewhere. It's just our
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resolution is not good enough to spot it. 500
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is the same deal. Okay, this one's a
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little quite a bit below 500 actually, a little
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above zero. Maybe around 200. You see slopes
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going up and then right below them going down,
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they have to be flat in the middle. Okay
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. So even though it's not as easy to see
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the straight lines on this one, this direction field
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does show to equilibrium solutions. So we've got a
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little less than 20. How about a little above
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? Okay, a little above 20. Now,
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this one's very different, right? We're basically pointing
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down everywhere. In fact, it's so vertical.
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It's hard to even tell that it's down. But
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it is down, okay? And we can be
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pretty confident. I mean, we never even get
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close to flat that there are no equilibrium solutions here
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. Okay? So for 10th and 20 there were
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there were equilibrium solutions for 30. There weren't.
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Um I should I should admit, with the benefit
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of hindsight, I know that the cut off should
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be a 20. But this is the kind of
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experimentation you want to do in order to determine that
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fact. Crazy. No. Let's get to the
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math. That tells us we're right about this.
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So I'm going to write down our differential equation.
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This is equal, Teoh. I'm switching to fractions
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because I think those air almost always easier to deal
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with who I forgot. Api. I'm gonna use
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some exponents here. Scientific notation. This is 10
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to the three. Instead of writing 1000 that's gonna
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be useful in a little bit. Minus c were
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interested in equilibrium, solutions and equilibrium solutions occur when
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the derivative is equal to zero. In this problem
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, we're gonna set our expression this middle one equal
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to zero. Because that is what it tells us
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. The magnitude of the derivative. And we noticed
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. At least it's it's beneficial to notice that,
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um Who I made a mistake here. Sorry about
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that. This is one and then this is he
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Okay, right? That looks better now. Sorry
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, I got some things jumbled in my head.
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No will return to the problem. So we're setting
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this equal to zero, and it's really beneficial to
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notice that this expression you have for the derivative is
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, in fact, a quadratic. Okay. And
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if you just distribute and do a little rearranging,
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you actually see that what we end up with is
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a minus A over 10 to the five. Um
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, he squared, plus a over 100 p minus
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C equals 20 Okay. And the beneficial thing is
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, we can plug this into our formula, right
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? So the quadratic formula, if you imagine this
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being, like, if you're used to a X
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squared plus b x plus C equals+20 that's the
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formula you're used to well, minus 8/10 to the
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five. That's our A, um, a over
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100 that's RB. And then see, a little
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confusingly is minus C. Okay. And once Republic
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plugged that into the formula, we get negative.
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B lesser minus root. B squared is 64/10 to
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the four. That's 10,000 minus four A. C
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to the tube gives in the multiplication cancel out,
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and we actually end up with 32 over 10 to
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the five seen go and that's all over two A
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. All right? And now ask yourself basically what's
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important in this formula, Okay, because we don't
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necessarily need a lot of exact values for P.
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And this problem, that's not what we're asked.
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What we're asked are things like, When do we
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have no solutions versus solutions? Um, when do
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we have actual die off? Okay, so a
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lot of that is gonna have to do is just
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this part under the square root, right? If
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this part under the square root is greater than zero
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, well, then we have plus or minus so
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a positive number. We take the square root.
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That's another positive number plus or minus that we get
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to solutions. Okay, one for the plus one
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for the minus. If it's equal to zero.
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Well, plus or minus zero, that's the same
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thing. That's just one solution. Okay? And
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if it's less than zero, well, we can't
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. Can't take a square root of a negative number
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. We're dealing just with real numbers. We can't
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do that. So that's actually going to give us
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no solutions, which is entirely possible. Okay.
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No. When does this happen? Well, the
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cut off point that were interested in between, um
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two solution between having one solution and were more one
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or more solutions and no solutions is where this expression
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is greater than or equal to zero. Because that
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encompasses both the two solutions and the one solution case
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. Now, it turns out after you do a
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little bit of algebra or not even a well,
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you do a little manipulation that that gives you so
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you you add over and you end up getting see
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equal to 20. So if you add, if
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you add, is the cut off point. So
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you add this over Now you multiply both sides by
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10 to the fifth over 32 you basically end up
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getting see being less than or equal to 20 or
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20 being greater than or equal to see. This
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is the case where you have one or more solutions
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. No, the original OD is no longer here
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, but what this means kind of when you're interpreting
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the problem is that you shouldn't be catching more than
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20 fish a day. All right, so you're
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somewhere between, um what the what the math tells
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us is between minus infinity and 20. But in
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reality, you know, if you're thinking about the
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problem, you're not gonna be catching negative fits.
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That would mean putting them back into the ocean.
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So really, you're between zero and 20 catches a
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day, and you're good if you have more.
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If you catch more than 20 fish a day,
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that's when you're just going to decay off. Okay
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? And what's happening is, Well, that's the
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case we plot. It's the 30 right? There
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are no equilibrium. Solutions just goes down. Um
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, and if you want a little more evidence than
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the graphical approach, basically it could be described this
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way minus C. So eventually what's gonna happen is
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when the sea gets too small. Well, this
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is a parable on the left, right. It's
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minus a over 10 to the five p squared,
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plus a over 100 p minus. See, this
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whole thing's a proble. And because this coefficient is
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less than zero, it's gonna be a Kong cave
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down parabola. Got a little messy there. It's
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gonna be so it's gonna look something like this because
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that's first coefficient is less than zero. So if
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you shift, see down for enough, you might
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ask, OK, why does it go to zero
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? Why does it decrease instead of increasing? Well
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, it's because the parabola is pointed down and be
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no zeros. There must be, is kind of
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a gap here, and it's gonna be negative the
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whole time, okay? And that means the derivative
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is going to be negative the whole time, okay
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. And that corresponds to going down. So if
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this were a conclave up proble, the interpretation would
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be that you're actually always increases. There still wouldn't
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be any equilibrium when they're when the quadratic gives,
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you know solutions, but you be increasing. But
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in this case, we're concave down, so we're
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decreasing the whole time on. If you catch more
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than 20 fish a day, you will kill off
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the entire population