SOLUTION 1 Let's take the semicircle to be the upper half of the circle x² + y² = r² with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the x- axis as shown in the top figure. 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 y = Thus A = The domain of this function is 0 sxsr. Its derivative is 2(2 – 2x²) VR - x2 2x2 A' = 22 2 - 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 27 SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let 0 be the angle shown in the bottom figure. Then the area of the rectangle is A(0) = (2r cos(0))(r sin(0)) = r(2 sin(0) cos(0)) = r? sin(20). We know that sin(28) has a maximum value of 1 and it occurs when 28 = n/2. So A(8) has a maximum value of r2 and it occurs when e = t/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all.
SOLUTION 1 Let's take the semicircle to be the upper half of the circle x² + y² = r² with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the x- axis as shown in the top figure. 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 y = Thus A = The domain of this function is 0 sxsr. Its derivative is 2(2 – 2x²) VR - x2 2x2 A' = 22 2 - 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 27 SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let 0 be the angle shown in the bottom figure. Then the area of the rectangle is A(0) = (2r cos(0))(r sin(0)) = r(2 sin(0) cos(0)) = r? sin(20). We know that sin(28) has a maximum value of 1 and it occurs when 28 = n/2. So A(8) has a maximum value of r2 and it occurs when e = t/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all.
Elementary Geometry for College Students
6th Edition
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Daniel C. Alexander, Geralyn M. Koeberlein
Chapter10: Analytic Geometry
Section10.CT: Test
Problem 22CT
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Transcribed Image Text:SOLUTION 1 Let's take the semicircle to be the upper half of the circle x² + y² = r² with center the origin.
Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the x-
axis as shown in the top figure.
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 y =
Thus
A =
The domain of this function is 0 sxsr. Its derivative is
2(2 – 2x²)
VR - x2
2x2
A' = 22
2 - 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
27
SOLUTION 2 A simpler solution is possible if we think of using an angle as a variable. Let 0 be the angle
shown in the bottom figure. Then the area of the rectangle is
A(0) = (2r cos(0))(r sin(0)) = r(2 sin(0) cos(0)) = r? sin(20).
We know that sin(28) has a maximum value of 1 and it occurs when 28 = n/2. So A(8) has a maximum
value of r2 and it occurs when e = t/4.
Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at
all.
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