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Numerade Educator

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Problem 58 Hard Difficulty

What is the smallest possible area of the triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola $ y = 4 - x^2 $ at some point?

Answer

Area $=\frac{32 \sqrt{3}}{9}$

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Video Transcript

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