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 x? + y? = 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 = rsine r cos To eliminate y we use the fact that (x, y) lies on the circle x2 + y2 = and so y = Thus A = Video Example ) Tutorial Online Textbook The domain of this function is 0 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(-) = 2 V2 V2 SOLUTION 2 A simpler solution is possible 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(8)) = r(2sin(0)cos(8)) = r'sin(20) We know that sin(20) has a maximum value of 1 and it occurs when 20 = x/2. So A(0) has a maximum value of r and it occurs when e = A/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all.
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 x? + y? = 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 = rsine r cos To eliminate y we use the fact that (x, y) lies on the circle x2 + y2 = and so y = Thus A = Video Example ) Tutorial Online Textbook The domain of this function is 0 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(-) = 2 V2 V2 SOLUTION 2 A simpler solution is possible 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(8)) = r(2sin(0)cos(8)) = r'sin(20) We know that sin(20) has a maximum value of 1 and it occurs when 20 = x/2. So A(0) has a maximum value of r and it occurs when e = A/4. Notice that this trigonometric solution doesn't involve differentiation. In fact, we didn't need to use calculus at all.
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter7: Conic Sections And Quadratic Systems
Section7.1: The Circle And The Parabola
Problem 3SC
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Transcribed Image Text: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 x? + y? = 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 =
rsine
r cos
To eliminate y we use the fact that (x, y) lies on the circle x2 + y2 = and so y =
Thus A =
Video Example )
Tutorial
Online Textbook
The domain of this function is 0 <x< r. Its derivative is
2 (12 – 2x²)
212
A' = 2Vr2 – x² –
Vr2 –
which is 0 when 2x = r, that is, x =
(since x > 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(-) = 2
V2
V2
SOLUTION 2 A simpler solution is possible
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(8)) = r(2sin(0)cos(8)) = r'sin(20)
We know that sin(20) has a maximum value of 1 and it occurs when 20 = x/2. So A(0) has a maximum value of r and it occurs when e =
A/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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