More on mechanics

1. Feb 9, 2006

stunner5000pt

of the classical kind

All i need here is an explanation as why F is the way it is
A person wished to fence off a maximum area for hi8s dog by attachin a flexible fence of length L to the side of his house whose width is 2a. What should the slope of the fence be?
Assume L > 2a
well then the action
$$A[y] = \int_{-a}^{a} y(x) dx$$ is a maximum
subject to constant L where $$L = \int_{-a}^{a} \sqrt{1+y'^2} dx$$
If L[y,y'} = constant then $\delta L = 0$
and A[y,y'] = constant at its maximum
then
$$\delta A + \lambda \delta L = 0$$ where
$$\lambda = \frac{[A]}{[L]}$$
we also want that $$\delta \int_{-a}^{a} F(y,y',x) dx = 0$$
where $$F = y + \lambda \sqrt{1+y'^2}$$
thats the problem, Why is F the way it is ? Maybe the attached diagram helps...

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Last edited: Feb 9, 2006
2. Feb 10, 2006

stunner5000pt

can anyone help out?

is it because f takes on the shape of half a unit circle?

Or will any function do?

3. Feb 11, 2006

lightgrav

maximizing the Area provided the "y" term,
using all the fence material provided the sqrt term.
Whether you put the lambda multplier with the sqrt term
or with the y term was your choice, back in step 4 ...