Maximising area for constant perimeter

In summary, the homework equation for a curve y = y(x) that meets the x-axis at x = a, x = -a, and has fixed length pi*a between these points is y(a) = y(-a) = 0. The equation of a circle that encloses the maximum area between itself and the x-axis is (x + B)^2 + (y + D)^2 = C^2.
  • #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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  • #2
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
(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
Are you sure it shouldn't be

lambda*(1 + (dy/dx)^2)^(1/2) + y ?
 
  • #5
SunGod87 said:
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
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
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
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!
 

What is the concept of maximising area for constant perimeter?

The concept of maximising area for constant perimeter is a mathematical problem that involves finding the largest possible area for a given perimeter. This is often encountered in real-world situations where there are limited resources or materials available.

Why is maximising area for constant perimeter important?

Maximising area for constant perimeter is important because it allows us to efficiently use resources and materials to achieve the largest possible area. This can have practical applications in fields such as architecture, agriculture, and manufacturing.

What is the formula for calculating the maximum area for a given perimeter?

The formula for calculating the maximum area for a given perimeter is A = (P/4)^2, where A is the area and P is the perimeter. This formula is derived from the fact that a square has the largest area for a given perimeter.

What are some strategies for maximising area for constant perimeter?

Some strategies for maximising area for constant perimeter include using shapes with equal sides, such as squares or equilateral triangles, as they have the largest possible area for a given perimeter. Additionally, dividing the perimeter into smaller segments and rearranging them to form a larger shape can also help maximise the area.

Are there any limitations to maximising area for constant perimeter?

Yes, there are limitations to maximising area for constant perimeter. This problem assumes that the shape being used is continuous and does not account for any gaps or spaces between shapes. It also does not consider the structural integrity or stability of the shape being used to maximise the area.

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