• Rectangle is of maximum perimeter.
• The rectangle is inscribed inside a circle.
• The radius of the circle is ‘a’.
Let us consider,
• ‘x’ and ‘y’ be the length and breadth of the given rectangle.
• Diagonal AC2 = AB2 + BC2 is given by 4a2 = x2+y2 (as AC = 2a)
• Perimeter of the rectangle, P = 2(x+y)
Consider the diagonal,
4a2 = x2 + y2
y2 = 4a2 – x2
Now, perimeter of the rectangle, P
P = 2x + 2y
Substituting (1) in the perimeter of the rectangle.
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.
[squaring on both sides]
4a2 – x2 = x2
2x2 = 4a2
x2 = 2a2
[as x cannot be negative]
Now to check if this critical point will determine the maximum diagonal, we need to check with second differential which needs to be negative.
Consider differentiating the equation (3) with x:
[Since and and if u and v are two functions of x, then ]
Now, consider the value of
As , so the function P is maximum at .
Now substituting in equation (1):
As the sides of the taken rectangle are equal, we can clearly say that a rectangle with maximum perimeter which is inscribed inside a circle of radius ‘a’ is a square.
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