Questions: The figure above shows a rectangle inscribed in a semicircle with a radius of 20. The area of such a rectangle is given by A(x)=2x√(400-x^2), where the width of the rectangle is 2x. It can be shown that A'(x)=-2x^2/√(400-x^2)+2√(400-x^2) and A has critical values of -20, -10√2, 10√2, and 20. It can also be shown that A'(x) changes from positive to negative at x=10√2. Which of the following statements is true? (A) The inscribed rectangle with maximum area has dimensions 10√2 by 10√2. (B) The inscribed rectangle with minimum area has dimensions 10√2 by 10√2. (C) The inscribed rectangle with maximum area has dimensions 20√2 by 10√2.

The problem states that the area of the rectangle is given by \(A(x) = 2x\sqrt{400 - x^2}\), where \(2x\) is the width of the rectangle. The problem also provides the first derivative of \(A(x)\) and states that the area function has a critical point at \(x = 10\sqrt{2}\) where \(A'(x)\) changes from positive to negative. We need to determine which statement about the dimensions of the inscribed rectangle with maximum area is true.

Find the width of the rectangle. The width is given as \(2x\). Since \(A(x)\) reaches a maximum at \(x = 10\sqrt{2}\), the width is \(2(10\sqrt{2}) = 20\sqrt{2}\).

Find the height of the rectangle. The height of the rectangle is given by \(\sqrt{400 - x^2}\). Substituting \(x = 10\sqrt{2}\), we get the height as \(\sqrt{400 - (10\sqrt{2})^2} = \sqrt{400 - 200} = \sqrt{200} = 10\sqrt{2}\).

Determine the dimensions of the rectangle with maximum area. The width of the rectangle is \(20\sqrt{2}\) and the height is \(10\sqrt{2}\). Thus, the dimensions of the inscribed rectangle with maximum area are \(20\sqrt{2}\) by \(10\sqrt{2}\).

\(\boxed{\text{(C)}}\)

\(\boxed{\text{(C)}}\)