Maximising area for constant perimeter

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  • #1
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Homework Statement


A curve y = y(x) meets the x-axis at x = a, x = -a and has fixed length pi*a between these points. Show that the curve which encloses the maximum area between itself and the x-axis is the semi circle x^2 + y^2 = a^2 for y => 0


Homework Equations


INT[-a,a] ds = INT[-a,a] (1 + (dy/dx)^2)^(1/2) = pi*a
y(a) = y(-a) = 0
 
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Answers and Replies

  • #2
Dick
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Apply the Euler-Lagrange equations to the unconstrained functional:
(1 + (dy/dx)^2)^(1/2)+lambda*y.
The lambda is a Lagrange multiplier representing your constraint and you solve for it in the end.
 
  • #3
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(1 + (dy/dx)^2)^(1/2) + lambda*y = F + lambda*G

lambda = C
dy/dx = y'

No explicit dependence on x so:
(1 + (y')^2)^(1/2) + C*y - (y')^2 (1 + (y')^2)^(-1/2) = A (constant)
(1 + (y')^2) + C*y*(1 + (y')^2)^(1/2) - (y')^2 = A*(1 + (y')^2)^(1/2)
1 + C*y*(1 + (y')^2)^(1/2) = A*(1 + (y')^2)^(1/2)
1 = (A - C*y)*(1 + (y')^2)^(1/2)
1/(A - C*y) = (1 + (y')^2)^(1/2)
1/(A - C*y)^2 = (1 + (y')^2)
(y')^2 = (1/(A - C*y)^2) - 1
y' = [(1/(A - C*y)^2) - 1]^(1/2)

Substitution: 1/(A - Cy) = coshz
y = A/C - sechz/C
dy = -(tanhz.sechz)/C

Making substitutions and seperating variables:
-1/C INT[sech^2 zdz] = x + B
tanh^2 z = C^2*(x + B)^2
sech^2 z = 1 - C^2*(x + B)^2
coshz = [1 - C^2*(x + B)^2]^(-1/2)
A - Cy = [1 - C^2*(x + B)^2]^(1/2)
Cy = A - [1 - C^2*(x + B)^2]^(1/2)
y = A/C - 1/C*[1 - C^2*(x + B)^2]^(1/2)
y = A/C - [(1/C^2) - (x + B)^2]^(1/2)

Is this okay so far? I then tried:
y(a) = y(-a) = 0, which gives B = 0
Similarly I can obtain A/C = (1/C^2 - a^2)^(1/2)

But then I run out of ideas, is this right so far?
 
  • #4
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Are you sure it shouldn't be

lambda*(1 + (dy/dx)^2)^(1/2) + y ?
 
  • #5
Dick
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Are you sure it shouldn't be

lambda*(1 + (dy/dx)^2)^(1/2) + y ?
In this case it doesn't matter much which you consider to be the constraint. Just changes lambda->1/lambda. Your solution looks to be a bit more complicated than it needs to be. But when you get down to here:

y = A/C - [(1/C^2) - (x + B)^2]^(1/2)

you are basically done. Rearrange and rename constants and it is (y+A)^2+(x+B)^2=R^2, a general equation for a circle. Now apply the contraints to determine A,B and R.
 
  • #6
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I tried another way and got the equation of a circle!
But I'm still stuck!

I have (x + B)^2 + (y + D)^2 = C^2
y(a) = y(-a) = 0
(a + B)^2 + (y + D)^2 = C^2
(-a + B)^2 + (y + D)^2 = C^2
so: (a + B)^2 = (-a + B)^2 and B = 0

x^2 + (y + D)^2 = C^2
Using the constraints:
y(a) = y(-a) = 0
INT[-a,a] (1 + (dy/dx)^2)^(1/2) dx = pi*a

How can I show that D = 0 and C = a, as the question requires?
 
  • #7
Dick
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From this point, think more geometrically. You KNOW it's circle. What circle arc connects the two points, has a center on the y-axis and a length of a*pi? Going back through the arc length formalism is needlessly painful.
 
  • #8
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Ah I can just write it like that. It makes sense but I was worried perhaps the lecturer was looking for something more. Thanks very much!
 

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