Expert: Scotto - 3/31/2011

Question
I can't seem to figure out how to solve this equation...I know I need to remove some of the variables but I dont know how to do this with no numbers.

Find the dimensions of the rectangle of maximum area that can be inscribed in a semicircle of radius a, if two vertices lie on the diameter.

Answer
The radius of the circle is a, so the diameter is 2a.
If the sides of the rectangle are b and c, then b^2 + c^2 = 4a^2.
This says that if if b=x, c=sqrt(4a^2-x^2).

Note that in  this triangle, it is not like normal with sides a and b,
but here the sides are b and c with the hypotenuse as 2a.

Since the area of the rectangle is bc, that means the area is f(x) = x*sqrt(4a^2 - x^2).

To find that maximum of f(x) = x*sqrt(4a^2 - x^2), take the derivative.
This can be written as f(x) = p(x)q(x) where p(x) = x and q(x) = sqrt(4a^2 - x^2).

It can be seen that p'(x) = 1 and q'(x) = -x/(sqrt(4a^2 - x^2) { there was a 2/2 involved }.

The way to finish this problem, then, is to solve p(x)q'(x) + p'(x)q(x) = 0 for x.
Since p(x) = x, this is the same as xq'(x) + q(x) = 0.
That is -x^2 / sqrt(4a^2 - x^2) + sqrt(4a^2 - x^2) = 0.

Multiply by sqrt(4a^2 - x^2) gives -x^2 + 4a^2 - x^2 = 0.
That simplifies to 2x^2 = 4a^2, so x^2 = 2a^2, so x = a*sqrt(2).

Since we had b=x and c=sqrt(4a^2-x^2), that gives us b = a*sqrt(2) and
c = sqrt(4a^2 - 2a^2) = sqrt(2a^2) = a*sqrt(2).  As can be seen, b=c,
and those were the two sides of this rectangle, so it is a square.