Search Textbook Solutior x 526408 Q2 Notes Ask Your Teach EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a s emicircle of radlusr SOLUTION 1 Let's take the semicircle to be the upper half of the circle x+y Then the word inscribed means that the rectangle has two vertices on the axis as shown in the top figure. with center the semicircle and two vertices on the x- Let (x, y) be the vertex that les in the first quadrant. Then the rectangle has sides of lengths 2x and y, so its area is A = To eliminate y we use the fact that (x, y) lles on the circle x2+ y2 2 and so y . Thus The domain of this function is 0 s x s r. Its derivative is 2x22-2x2 which is 0 when 2x2=r2, that is, x = (since x 2 0). This value of x gives a maximum value of A since A(O)-0 and A(r) 0. Therefore the area of the largest inscribed rectangle is SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let 8 be the angle shown in the bottom figure. Then the area of the rectangle is A(0)-(2r cos(θ))(r sin(θ))-r2(2 sin(θ) cos(θ)) = r2 sin(28). We know that sin(20) has a maximum value of 1 and it occurs when 2θ = Ï/2. So A(9) has a maximum value of r2 and it occurs when θ = Ï/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all. ^ d, 7:30 PM 11/26/2017