Prove that the largest rectangle with a given perimeter 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.
• Area of the rectangle, A = x × y.
Now as consider the perimeter of the rectangle,
p = 2(x +y)
p = 2x + 2y
----- (1)
Consider the area of the rectangle,
A = x × y
Substituting (1) in the area of the rectangle,
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----- (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:
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[Since
]
----- (3)
To find the critical point, we need to equate equation (3) to zero.
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Now to check if this critical point will determine the largest rectangle, we need to check with second differential which needs to be negative.
Consider differentiating the equation (3) with x:
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[Since
and
]
------ (4)
Now, consider the value of ![]()
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As
, so the function P is maximum at
.
Now substituting
in equation (1):


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