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

Mar 20, 2017

Let us set up the following variables:

$\left\{\begin{matrix}x & \text{Width of the rectangle" \\ y & "Height of the rectangle" \\ P & "Perimeter of the rectangle}\end{matrix}\right.$

Our aim is to find $P \left(x\right)$, (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 $\frac{\mathrm{dP}}{\mathrm{dx}}$.

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

$\setminus \setminus \setminus {A}^{2} = x y$
$\therefore y = {A}^{2} / x$ ..... [1]

And the total Perimeter of the rectangle is given by:

$P = x + x + y + y$
$\setminus \setminus \setminus = 2 x + 2 y$

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

$P = 2 x + \frac{2 {A}^{2}}{x}$ ..... [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:

$\frac{\mathrm{dP}}{\mathrm{dx}} = 2 - \frac{2 {A}^{2}}{x} ^ 2$

At a critical point we have $\frac{\mathrm{dP}}{\mathrm{dx}} = 0 \implies$

$2 - \frac{2 {A}^{2}}{x} ^ 2 = 0$
$\therefore \setminus \setminus \setminus {A}^{2} / {x}^{2} = 1$
$\therefore \setminus \setminus \setminus \setminus \setminus {x}^{2} = {A}^{2}$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus x = \pm A$
$\therefore \setminus \setminus \setminus \setminus \setminus \setminus \setminus x = A \setminus \setminus \setminus \because x > 0 , \text{ and } A > 0$

Hence we have $x = A \implies y = A$ (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:

$\frac{{d}^{2} P}{\mathrm{dx}} ^ 2 = \frac{4 {A}^{2}}{x} ^ 3 > 0 \text{ when } x = A$

Confirming that we have a minimum perimeter

QED

Mar 20, 2017

See explanation below.

#### Explanation:

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

$p = 2 \left(x + y\right)$

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

$x y = {A}^{2}$ or $x y - {A}^{2} = 0$

We then form the Lagrange function:

$\Lambda \left(x , y , \lambda\right) = 2 \left(x + y\right) + \lambda \left(x y - {A}^{2}\right)$

and equate the gradient of $\Lambda$ to zero:

$\frac{\partial \Lambda}{\partial x} = 2 + \lambda y = 0$

$\frac{\partial \Lambda}{\partial y} = 2 + \lambda x = 0$

$\frac{\partial \Lambda}{\partial \lambda} = x y - {A}^{2} = 0$

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

$p = 4 A$

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

$x = A + \mathrm{dx}$

$y = {A}^{2} / x = {A}^{2} / \left(A + \mathrm{dx}\right)$

Then:

$p = 2 \left(A + \mathrm{dx} + {A}^{2} / \left(A + \mathrm{dx}\right)\right) = 2 \frac{{\left(A + \mathrm{dx}\right)}^{2} + {A}^{2}}{A + \mathrm{dx}} = 2 \frac{2 {A}^{2} + 2 A \mathrm{dx} + {\mathrm{dx}}^{2}}{A + \mathrm{dx}} = \frac{4 A \left(A + 2 x\right) + 2 {\mathrm{dx}}^{2}}{A + \mathrm{dx}} = 4 A + \frac{2 {\mathrm{dx}}^{2}}{A + \mathrm{dx}} > 4 A$