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.


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