EXAMPLE 5 Find the area of the largest rectangle that can be inscribed in a semicircle of radius r. SOLUTION 1 Let's take the semicircle to be the upper half of the circle x2 + y2 =2 with (x, y) center the origin. Then the word inscribed means that the rectangle has two vertices on the 2.x semicircle and two vertices on the x-axis as shown in the top figure. 0 Let (x, y) be the vertex that lies 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) lies on the circle x2 + y2 =r2 and so r sin 8 y = Thus A = r cos 0 The domain of this function is Os xsr. Its derivative is Video Example () 2x2 A' = 2V12 - x2 2(12 - 2x 2 ) V12 - x2 V12 - x2 which is 0 when 2x2 = /2, that is, x = (since x 2 0). This value of x gives a maximum value of A since A(0) = 0 and A(r) = 0. Therefore the area of the largest inscribed rectangle is A ( VZ ) = 2 VZ SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let @ be the angle shown in the bottom figure. Then the area of the rectangle is A(0) = (2r cos(0))(r sin(0)) = r2(2 sin(@) cos(0)) = 12 sin(20). We know that sin(20) has a maximum value of 1 and it occurs when 20 = 71/2. So A(0) has a maximum value of / and it occurs when 0 = 7/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all. Need Help? Read It