Maximising Area (1 Viewer)

[Fus Ro Dah]

Consider a straight line and a piece of string with length L.

In what shape should the string be made to maximise the area bounded by the straight line and the string, given that the string must touch the line at least once?

Give proof.

 

[Carrotsticks]

I have an idea for a proof. Will post when I get home because I will need my trusty tablet

But just for fun, define the set of points bounded by the string and the line to be S such that whatever it doesn't cover is S'.

Then define S to be the outside!

EDIT: Home now

Here is a diagram:

Define a reflective bijection about the given line which maps as shown below (sorry it looks a bit wonky. Should be an invariant transformation):

So now we have the 'perimeter' of this closed set to be 2L.

By the Isoperimetric Inequality, the area of the region is bounded by:



Where A denotes the area of the closed region. Note that this is a slight variation of the actual inequality since our perimeter is length 2L not L.

But since we want maximal area, we take the supremum:



And so this is our maximum area. We now prove that it is in fact a circle.

Suppose we have a circle with circumference length 2L.



But remember that this is the area of and so since C' is just a (area preserving) bijection of C, we have C = C'

And so therefore our maximum area is:



------------------------------------------------------------------------------

However, I'm not 100% sure how to do this using 4U methods.

My guess would be go construct an arbitrary region (sort of like the use of polar coordinates in evaluating area) and partition it as such:

And sum the areas of the triangles and using Calculus, show that max{A} occurs when the partition lengths are equal in length. Problem is, this doesn't necessarily imply it's a semi-circle...

 

[Carrotsticks]

Oh woops I forgot that the inequality I provided above only works in of Euclidean Space.

So scratch the (x_1, x_2, x_3, ... , x_n) --> (x_1 ', x_2 ', x_3 ', ... , x_n ')

So it's the two-variable case: (x_1, x_2) --> (x_1 ', x_2 ')

 

[iSplicer]

Carrot, you hero!

 

[seanieg89]

I have an idea for a proof. Will post when I get home because I will need my trusty tablet

But just for fun, define the set of points bounded by the string and the line to be S such that whatever it doesn't cover is S'.

Then define S to be the outside!

EDIT: Home now

Here is a diagram:

Define a reflective bijection about the given line which maps as shown below (sorry it looks a bit wonky. Should be an invariant transformation):

So now we have the 'perimeter' of this closed set to be 2L.

By the Isoperimetric Inequality, the area of the region is bounded by:



Where A denotes the area of the closed region. Note that this is a slight variation of the actual inequality since our perimeter is length 2L not L.

But since we want maximal area, we take the supremum:



And so this is our maximum area. We now prove that it is in fact a circle.

Suppose we have a circle with circumference length 2L.



But remember that this is the area of and so since C' is just a (area preserving) bijection of C, we have C = C'

And so therefore our maximum area is:



------------------------------------------------------------------------------

However, I'm not 100% sure how to do this using 4U methods.

My guess would be go construct an arbitrary region (sort of like the use of polar coordinates in evaluating area) and partition it as such:

And sum the areas of the triangles and using Calculus, show that max{A} occurs when the partition lengths are equal in length. Problem is, this doesn't necessarily imply it's a semi-circle...

Isn't the isoperimetric inequality a problem of essentially equal difficulty? By assuming it it seems we are circumventing the main bulk of the problem...Also, showing that a circle in particular attains the bound that the isoperimetric inequality asserts does not guarantee that ANY such maximal shape must be a circle.

Regarding the attempt using triangles and calculus, some geometric work must be done first, as our diagram gets a whole lot more messy if the region is not convex. Perhaps the problem could be phrased within the realms of 4-unit by imposing lots of regularity conditions on the curve/region?

To prove the general isoperimetric inequality I think you probably need something pretty powerful, like Green's theorem or calculus of variations.

EDIT: It turns out there are some rather clever elementary proofs that any solution that exists must be a circle, but it is a little harder to prove existence.

 

[Carrotsticks]

To prove the general isoperimetric inequality I think you probably need something pretty powerful, like Green's theorem or calculus of variations.

Yep I was actually taking a look at that a couple weeks ago (upon learning Green's Theorem).

A combination of Green's Theorem and the Cauchy-Schwarz inequality yields the isoperimetric inequality as given above, but not quite sure about the generalised inequality.

I have a feeling OP just thought of it, rather than obtaining it from an actual textbook. Waiting for his reply now.

 

[seanieg89]

Yep. Alas Green's theorem is actually not so easy to prove properly... I have been studying vector calculus/calculus on manifolds again for the last couple of weeks, because I realised how little I actually understood the big theorems in undergrad (beyond being able to blindly apply them in simple situations).

 

[Carrotsticks]

Barbernator, I saw that =p

Anyway, I think this thread is best moved to extracurricular.

 

[Fus Ro Dah]

I have a feeling OP just thought of it, rather than obtaining it from an actual textbook. Waiting for his reply now.

Yeah I just thought of it randomly. Sorry if it was too much trouble. I honestly expected a more elementary solution easily found by the average MX2 student =/