## Solutions

Let ABCD be rectangle which is  inscribed in a given circle of radius ‘r’ And Let θ be the angle between side of rectangle and Diameter of given circle.

Therefore from right angled  Δ ABC ,

We have AB  = AC cosθ          ∵ AC = 2r
Let A(x) be the area of Rectangle ABCD
∴ A(x) = AB × BC
A = (2r cos θ)(2r sin θ )
A =  4r2 sin θ cos θ
A = 2r2  (2sin θ cos θ)
A = 2r2  (sin 2θ )

⇒ 2r2 2 (cos 2θ ) = 0 ,As r2 is constant
⇒cos 2θ = 0
⇒cos 2θ =cos (π/2)
⇒ θ = π/4

∴ A has Maximum value at θ = π/4

## Solution

Let us consider two numbers x and 16- x .
Then transforming our problem to mathematical form which says “sum of whose cube”  as follows
A (x) =   x3 + (16 - x)3…….. (1)
Differentiating both sides w .r. t  “x” , we get

X = 8
So  x  =  8 will be the 1st required numbers if Double derivatives of A  w. r. t  ‘x’ comes to be positive at x = 8.
Differentiate (2)  w. r. t. ‘x’  .

## FINAL WORDS

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