Fox 350

Based on the evidence presented by Bleaug, this is updated as follows:

Thank you.

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Fox 312 - Solutions

Two solutions for Fox 312 are identified below.

Bleaug:
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).

Migue:
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.

Fox 312

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Fox 257 - Solution

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See Fox 257.

Fox 264

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Die no soon, before all your story are told.
--- Yoda Fox

Fox 263

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One more before releasing a general case.

Fox 257

Let's build upon this one...
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