## Sunday, October 31, 2010

### Fox 312

1. This comment has been removed by the author.

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.

MIGUE.

5. This comment has been removed by the author.

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.