Sunday, October 31, 2010

Fox 312


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  2. Newzad,
    Your solution doens't look right. If x is given as in your figure then, the slope of the tangent line is NOT -f'(x). It is something different. You may want to check your solution.

  3. Yes it is wrong, thank you

  4. If area is A the answer is f(x)=y=A/2x.
    f(x) is the envelope of all straight lines that close in with axes X-Y constant area equals to A.
    Equation of one of this lines is: x/a + y/b = 1
    (a,0) in X & (0,b) in Y. Notice that a & b are two variable parameters.
    Another equation is: ab/2 = A (area of triangle that straight line draws with axes).
    (1) Let f(x,y,a,b)= bx+ay-ab = 0
    (2) Let g(a,b)=ab-2A = 0
    (3) f'a · g'b - f'b · g'a = 0 (Jacobian of derivatives respect to a and b parameters).
    f'a is de derivative of f respect a, ...

    y/b - x/a = 0 (3)

    If we resolve this system of equations (1),(2) & (3) we obtain:
    a = 2x & b = 2y -> ab = 4xy = 2A -> xy = A/2 (equilateral hyperbola) and y = f(x) = A/2x.

    Answer is D) for A =1.


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  6. Hello, it's been a while since I've been on here but I plan to start on these problems again since they are really good. Anyways, here is my answer.

    slope of the tangent line is just -y/x. We also know that yx = 1. Substitute in for y and you get -2/x^2, which is the derivative of the function. Take the integral and you get 2/x + C, and let C = 0. Therefore, f(x) = 2/x.

  7. Newzad, six,
    Your answers are identical. But if f(x)=2/x, then Area A remains constant at 4. So, 2/x can't be the answer. Do you see any problem in Bleaug's or Migue's solutions?

    Welcome back six!

  8. How does he get y = A/2x? Shouldn't it be y = 2A/x? If the area of the triangle is yx/2 = 1, then shouldn't it be y = 2/x? I believe this is where the problem lies. I may be seeing this incorrectly though.

  9. Nevermind, I read it wrong. His argument looks correct to me.