Tuesday, November 16, 2010

Fox 312 - Solutions

Two solutions for Fox 312 are identified below.

As in Fox 262 (see its solution here) A area (= OQR area below) extremal/constant implies P midpoint of QR. Because:
A constant <=> dA = 0 <=> area(PRR') = area(PQQ') <=> area(P'RR') = area(P'QQ') => PR = PQ when dP tends towards 0.

Hence A = 1 = 2xy or y = 1/2x i.e. answer (D).

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 the 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.


  1. This comment has been removed by the author.

  2. where do you read that "derivative of f is straight line"?

    if you think the answer must be a parabola, just consider asymptotic behaviour of the curve we are looking for: obviously both Ox and Oy axes are asymptotes. This is not compatible with a parabola.

    if still in doubt, do the reverse problem: compute area A of triangle defined by tangent to y=1/2x curve and 0x + Oy axes.


  3. yes you are right. it was a mistake. I forgot to delete the comment. Also I have a geometric proof, which does not need calculus. I will publish on my blog.

  4. http://geometri-problemleri.blogspot.com.es/2012/09/problem-111-ve-cozumu.html