# 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 ...