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)
. (using the hypotenuse formula)

Now as consider the perimeter of the rectangle,
p = 2(x +y)
p = 2x + 2y
 ----- (1)
 ----- (1)
Consider the diagonal of the rectangle,

Substituting (1) in the diagonal of the rectangle,

[squaring both sides]
 ----- (2)
 ----- (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)
 ------ (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
 and  ]
 ]
 ------ (4)
 ------ (4)
Now, consider the value of 

As  , so the function Z is minimum at
, so the function Z is minimum at  .
.
Now substituting  in equation (1):
 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.
 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.