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.

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.

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.

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?

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.

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ReplyDeleteNewzad,

ReplyDeleteYour 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.

Yes it is wrong, thank you

ReplyDeleteIf area is A the answer is f(x)=y=A/2x.

ReplyDeletef(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.

This comment has been removed by a blog administrator.

ReplyDeleteHello, 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.

ReplyDeleteslope 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.

Newzad, six,

ReplyDeleteYour 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!

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.

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

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

ReplyDelete