Question (Applications of Differentiation)

let w be the width of the rectangle
then the length is √[(2a)^2 -w^2]
=√(4a^2 -w^2)
area of the rectangle
A= w＊√(4a^2 -w^2)

Differentiate w.r.t. w
dA/dw = √(4a^2 -w^2) + (1/2)(-2w^2)/√(4a^2 -w^2)
= √(4a^2 -w^2) + (-w^2)/√(4a^2 -w^2)
=[ (4a^2 -w^2 -w^2)]/√(4a^2 -w^2)
=[ (4a^2 -2w^2)]/√(4a^2 -w^2)

put dA/dw = 0
[ (4a^2 -2w^2)]/√(4a^2 -w^2) =0
4a^2 -2w^2=0
2w^2 =4a^2
w =(√2)a

length =√(4a^2 -w^2)
=(√2)a

area =(√2)a＊(√2)a
=2a^2

Let the rectangle be ABCD and angle DBC be x. Since radius = a, so BD = 2a.
Therefore, DC = 2a sinx and BC = 2a cosx.
Area of rectangle, A = DC x BC = (2asinx)(2acosx) = 2a^2sin2x.
dA/dx = 4a^2 cos 2x.
Put it = 0, we get cos 2x = 0
2x = 90
x = 45.
Therefore, max. area happens when x = 45 degree.
So max area = 2a^2 sin 90 = 2a^2.