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

##### 2 Answers

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

Now, the total areas is given as

# \ \ \ 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,

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

At a critical point we have

# 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 **a square of side #A#**

We should check that

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

Confirming that we have a minimum perimeter

QED

See explanation below.

#### Explanation:

Consider a rectangle of sides

so we want to minimize

We then form the Lagrange function:

and equate the gradient of

Combining the first two we have:

which needs to hold for any

then from the third we have:

and

This is certainly a stationary point for

Then: