What is the maximum area of rectangle that can be inscribed in a circle?

According to the question,

The diagonal of the rectangle = Diameter of the circle.

√(L2 + B2) = 4   

⇒ L2 + B2 = 42    ----(1)

For maximum area of rectangle 

Length should be equal to it breadth [ L = B]

So, from (1) 

L = B = √8

∴ Area of rectangle = L × B = √8 × √8 = 8

Detailed Mehtod

Formula used:

Area of rectangle = length × breadth

Diameter of circle = 2 × Radius of circle 

\(\rm \frac{d}{dx} x^{n} = nx^{n - 1}\)

Calculation:

Let the length of the rectangle be L and its breadth be B

Since the rectangle is inscribed inside the circle, its diagonal has to be the diameter of the circle.

√(L2 + B2) = 4   

⇒ L2 + B2 = 42    ----(1)

The area of the rectangle is given by Lb

Differentiating the condition w.r.t L,

2L + 2b \(\rm \frac{dB}{dL}\) = 0

⇒ \(\rm \frac{dB}{dL} = \frac{-L}{B}\)       ----(2)

Now, differentiating the area w.r.t l since we need to maximize it,

B + L \(\rm \frac{dB}{dL}\) = 0       ----(3)

Substituting equation (2) into equation (3), we have

⇒ B + L \(\rm \frac{-L}{B}\) = 0   

⇒ B2 = L2

⇒ B = L

Substituting this relation in equation (1), we get

⇒ 2L2 = 42 or L2 = 8

⇒ area = LB = L2 = 8 square units

∴ The maximum area of rectangle inscribed in the circle is 8 sq.units.