Question

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

Answer 1

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, D = √x²+y². (using the hypotenuse formula)

Now as consider the perimeter of the rectangle,

p = 2(x +y)

p = 2x + 2y

y = p - 2x/2 ----- (1)

Consider the diagonal of the rectangle,

D = √x²+y²

Substituting (1) in the diagonal of the rectangle,

D = \(\sqrt{x^2+(\frac{p-2x}{2})^2}\)

[squaring both sides]

Z = D² = x²+(p - 2x/2)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:

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

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:

Now, consider the value of

so the function Z is minimum at x = p/4

Now substituting x = p/4 in equation (1):

As x = y = p/4 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.