Re: Optimizing Rectangle for Perimeter

As a loose proof, this works, but I think it's a little messy to bring
in 100 "units" unless it's strictly required of you. Best to deal with
the matter in "pure" abstract terms in proofs and talk about P, x and y
in terms of A.

Let me try a stricter version:

A = xy; A is constant and > 0. (1)
P = |2x + 2y|. (2) (Note: modulus notation used because we always want
P > 0)

(1) => y = A/x

(2) => P = 2x + 2A/x
=> P' = 2 - 2A/x^2

(Note: P' = 0 at critical point.)

=> 2 - 2A/x^2 = 0
=> x^2 = A
=> x = +/- sqrt(A)

y = A/x = +/- sqrt(A)

Thus the solutions are the critical points (x,y) : (-sqrt(A), -sqrt(A))
and (sqrt(A), sqrt(A))

Both solutions give the same value for P:

P = |2x + 2y| = 4sqrt(A)

As pointed out by Paul Sperry, for a strict proof we still need to show
that 4sqrt(A) is an absolute minimum for P;

We can do this by splitting our domain into various spaces around our
critical points. We can do this by splitting the domain space into
different areas around the critical points and considering how they
satisfy equations (1) and (2).

Area 1: x < -sqrt(A) and y < -sqrt(A) -- "bottom left" corner
Area 2: -sqrt(A) < x < sqrt(A) and y < -sqrt(A) -- "bottom centre"
square
Area 3: x > sqrt(A) and y < -sqrt(A) -- "bottom right" corner
Area 4: x < -sqrt(A) and -sqrt(A) < y < sqrt(A) -- "middle left"
Area 5: -sqrt(A) < x < sqrt(A) and -sqrt(A) < y < sqrt(A) -- "centre"
Area 6: x > sqrt(A) and -sqrt(A) < y < sqrt(A) -- "middle right"
Area 7: x < -sqrt(A) and y > sqrt(A) -- "top left"
Area 8: -sqrt(A) < x < sqrt(A) and y > sqrt(A) -- "top centre"
Area 9: x > sqrt(A) and y > sqrt(A) -- "top right"

In each area, either the parameters x and y will not satisfy (1), or it
will give P > 4sqrt(A).

Thus our proof is concluded. (I think.)

.

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