WEBVTT
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were asked what The smallest possible area of the triangle
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that is cut off by the first quadrant and whose
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hypotenuse tangents to the problem of four equals y equals
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four minus X squared at some point is sort of
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a long question. Tell us. Well, first
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of all we have are parabola y equals four minus
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X squared. In the derivative is the slope of
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the tangent line to Wyatt X. So why prime
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negative two X. This is our slope at X
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. Mhm. Now the area of our triangle A
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is gonna be one half times the base b times
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the height where the base B is the distance from
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the origin of the X intercept and the height is
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the distance from the origin to the Y intercept.
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Now we know that, um, the tangent line
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to Y equals four minus X squared. At some
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point, let's say X equals p Well, this
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contains the point p four minus p squared. And
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so the tangent line is why minus four minus p
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squared. This is equal to the slope, which
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is negative two p times x minus p. Yes
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. Yeah. So solving for why we get bless
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you y equals negative two p x plus p squared
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. I want to not no, they were getting
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plus four. No, mhm, you're now we
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can find the y intercept by plugging in X equals
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zero. This is the point zero and then p
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squared plus four and the X intercept of this line
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we set y equals zero. This is negative p
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squared plus four over negative two p, which is
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p squared plus four over two p zero. Now
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we'll substitute these into our area equations. We can
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write our area equation as a function of P,
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so area as a function of P is one half
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times or base, which is p squared plus four
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over two p times the height, which is p
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squared plus four. Yeah, stirrings. All right
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. And this eventually simplifies to 1/4 p cubed plus
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two p plus four overpay. No. Now,
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in order to minimize this area, I want to
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take your derivatives is just got to do it.
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A prime of p equals. This is 3/4 p
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squared plus two minus four over p squared. Been
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saying and we want to find when this is equal
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to zero we solve this rational equation? Well,
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we can write this as three p to the fourth
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plus eight p squared minus 16. Wow. Yeah
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, Slender. So exports over four p squared strange
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equals zero in the numerator. We have a quadratic
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in p squared. We solve this. Using quadratic
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formula 19 7 p squared equals negative eight plus or
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minus the square root of B squared, which is
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64 minus four times a so four times, three
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times c, which is negative 16 all over two
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times three. And this simplifies to negative eight plus
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or minus squared of 256 over six. Then now
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P squared, of course, has to be positive
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. The only way this is positive is if we
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have negative eight, plus route 2 56. So
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we're only looking at negative eight plus 16/6, which
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is four thirds most. And therefore we have that
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p is equal to once again because we're in the
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first quadrant. P has to be positive. So
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P is the positive square root of four thirds,
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which is to over three fired. Quick. Now
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, of course, if you test a prime on
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intervals to either side of two of the Route three
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. You should find that a prime of P is
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less than zero for P less than two of route
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three. And a prime of P is greater than
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zero for P greater than two of the Route three
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. And by the first derivative test, it follows
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that A has a minimum. I at P equals
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two over route three. Shit plugging in the value
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of P to our area equation. We get that
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area at two of the Route three. This is
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our minimum area. This simplifies to 32 Route three
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over nine after a few steps, which is approximately
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6.15 84