Given the perimeter of a rectangle, show that its diagonal is minimum when it is a square.

Given,

• Rectangle with given perimeter.

Let us consider,

• ‘p’ as the fixed perimeter of the rectangle.

• ‘x’ and ‘y’ be the sides of the given rectangle.

• Diagonal of the rectangle, . (using the hypotenuse formula)

Now as consider the perimeter of the rectangle,

p = 2(x +y)

p = 2x + 2y

----- (1)

Consider the diagonal of the rectangle,

Substituting (1) in the diagonal of the rectangle,

[squaring both sides]

----- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with x and then equating it to zero. This is because if the function f(x) has a maximum/minimum at a point c then f’(c) = 0.

Differentiating the equation (2) with respect to x:

[Since ]

= 2x - p + 2x

------ (3)

To find the critical point, we need to equate equation (3) to zero.

4x –p = 0

4x = p

Now to check if this critical point will determine the minimum diagonal, we need to check with second differential which needs to be positive.

Consider differentiating the equation (3) with x:

= 4 + 0

[Since and ]

------ (4)

Now, consider the value of

As , so the function Z is minimum at .

Now substituting in equation (1):

As the sides of the taken rectangle are equal, we can clearly say that a rectangle with minimum diagonal which has a given perimeter is a square.