A rectangle is to be placed in a semicircle of radius 2. What can be the largest area of the rectangle and what are its dimensions.

Let AB be the diameter, and O be the centre of the circle
of radius diameter. To find the rectangle of
maximum area.

Let C and D be the points on either side of
the centere O from which CE and DF are drawn perpendicular to the diameter AB to meet
the circumference at E and F on one side and E' and F' on the other side of the
circle.  

Therefore , the rectangle CDFE has the area =
CD*CE.

But OCE is a right angled triangle with OC = x say.
And CE^2 = OE^2-x^2 = r^2-x^2 = 2^2 = x^2 = 4-x^2 as r = 2 is
given.

So CD = 2x and CE =
sqrt(4-x^2).

Therefore Area of the rectangle A(x) =
2x*sqrt(4-x^2).

A(x) is therefore maximum for x = c, where
c is a solution of A '(x) = 0 and A"(c) < 0.

A'(x) =
{xsqr(4-x^2)}' = (x)'sqr(4-x^2)
+x{sqrt(4-x^2)}'

A'(x)=1*sqrt(4-x^2) +
{x/2sqr(4-x^2)}(4-x^2)'

A'(x)=sqrt(4-x^2)+x(-2x)/2sqrt(4-x^2).

A'(x)=
{2(4-x^2) -2x^2)/2sqrt(4-x^2) = 4(2-x^2)/2sqrt(4-x^2)

A'(x)
= (2-x^2)/sqr(4-x^2).

A'(x) = 0 gives:  Numerator 2-x^2  =
0. Or x = sqrt 2.

A"(x) =  {2-x^2)/sqrt(4-x^2)}' = 
{-2x sqrt(4-x^2)  -(2-x^2) (-2x)/2sqrt(4-x^2)}/(4-x^2).

=
-2x{sqr(4-x^2)-(2-x^2)/(4-x^2)}/(4-x^2)

=  -2x{(4-x^2
-2+x^2}/(4-x^2)^(3/2)

=
-2x*2/(4-x^2)^(3/2)

A"(sqrt2) :  -2*2/(4-2)^3/2
<0.

So for x = sqrt2, the area of the recrangle A(x)
= 2x^2 = 2(sqrt2)^2 = 4sq units is the maximum area.