The area of a rectangle is `A^2`. Show that the perimeter is a minimum when it is square?

Let us set up the following variables:

`{(x, "Width of the rectangle"), (y, "Height of the rectangle"), (P, "Perimeter of the rectangle") :}`

Our aim is to find `P(x)`, (a function of a single variable) and to minimize P, wrt `x` (equally we could the same with `y` and we would get the same result). ie we want a critical point of `(dP)/dx`.

Now, the total areas is given as `A^2` (constant) and so:

`\ \ \ A^2=xy`
`:. y=A^2/x` ..... [1]

And the total Perimeter of the rectangle is given by:

`P = x+x+y+y`
`\ \ \ = 2x+2y`

And substitution of the first result [1] gives us:

`P = 2x+(2A^2)/x` ..... [2]

We no have achieved the first task of getting the perimeter, `P`, as a function of a single variable, so Differentiating wrt `x` we get:

`(dP)/dx = 2 - (2A^2)/x^2`

At a critical point we have `(dP)/dx=0 =>`

`2 - (2A^2)/x^2 = 0`
`:. \ \ \ A^2/x^2 = 1`
`:. \ \ \ \ \ x^2 = A^2`
`:. \ \ \ \ \ \ \x = +-A`
`:. \ \ \ \ \ \ \x = A \ \ \ because x>0, " and " A>0`

Hence we have `x=A => y=A` (from [1]), ie a square of side `A`

We should check that `x=A` results in a minimum perimeter. Differentiating [2] wrt x we get:

`(d^2P)/dx^2 = (4A^2)/x^3 > 0 " when " x=A`

Confirming that we have a minimum perimeter

QED

Consider a rectangle of sides `x` and `y`. The perimeter is:

`p = 2(x+y)`

so we want to minimize `p` subject to the constraint:

`xy=A^2` or `xy-A^2 = 0`

We then form the Lagrange function:

`Lambda(x,y,lambda) = 2(x+y)+lambda(xy-A^2)`

and equate the gradient of `Lambda` to zero:

`(del Lambda) /(del x) =2+lambday=0`

`(del Lambda) /(del y) =2+lambdax=0`

`(del Lambda) /(del lambda) =xy -A^2=0`

Combining the first two we have:

`2+lambda y = 2 + lambda x`

which needs to hold for any `lambda`, so it implies:

`x = y`

then from the third we have:

`x = y = A`

and

`p=4A`

This is certainly a stationary point for `p(x,y)`. To prove that it is a minimum we can proceed directly: suppose we have a rectangle with sides:

`x= A+dx`

`y = A^2/x = A^2/(A+dx)`

Then:

`p= 2(A+dx+A^2/(A+dx))= 2((A+dx)^2+A^2)/(A+dx) = 2(2A^2+2Adx+dx^2)/(A+dx) = (4A(A+2x)+2dx^2)/(A+dx) =4A +(2dx^2)/(A+dx) > 4A`