a rectangle is constructed with its base on the diameter of a semicircle with radius 6 and with its two…

Explanation:

Step1: Set up the coordinate - system

Place the semicircle centered at the origin with radius (r = 6). Its equation is (y=\sqrt{36 - x^{2}}) ((y\geq0)). Let the base of the rectangle lie on the (x) - axis and the vertices on the semicircle have coordinates ((x,y)) and ((-x,y)). The base of the rectangle (b = 2x) and the height (h=y=\sqrt{36 - x^{2}}), so the area of the rectangle (A=2x\sqrt{36 - x^{2}}), where (0\leq x\leq6).

Step2: Differentiate the area function

First, square the area function (A^{2}=4x^{2}(36 - x^{2})=144x^{2}-4x^{4}). Differentiate (u = A^{2}) with respect to (x): (u^\prime=\frac{d}{dx}(144x^{2}-4x^{4})=288x - 16x^{3}).

Step3: Find the critical points

Set (u^\prime = 0), so (288x - 16x^{3}=16x(18 - x^{2}) = 0). We have (x = 0), (x=\sqrt{18}=3\sqrt{2}), and (x =-\sqrt{18}) (but since (x\geq0) in our context, we discard (x =-\sqrt{18})). When (x = 0), (A = 0); when (x = 3\sqrt{2}), (y=\sqrt{36-(3\sqrt{2})^{2}}=\sqrt{36 - 18}=\sqrt{18}=3\sqrt{2}).

Step4: Determine the dimensions

The base (b = 2x), substituting (x = 3\sqrt{2}), we get (b = 6\sqrt{2}), and the height (h = 3\sqrt{2}).

Answer:

The rectangle with maximum area has base (6\sqrt{2}) and height (3\sqrt{2}).