rcos 9 (x, y) rsin @ 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 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 =
rcos 9 (x, y) rsin @ 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 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 =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:A
2x
r cos 9
Video Example
(x, y)
rsin 0
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 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 =
The domain of this function is 0 ≤ x ≤ r. Its derivative is
2x²
2 (²2²)
A' = 2√√²-2-
which is 0 when 2x² = r², that is, x =
(since x > 0). This value of x gives a maximum value of A
=
0 and A(r) = 0. Therefore the area of the largest inscribed
since A(0)
rectangle is
Thus A =
A(-
r
= 2
P
√2
SOLUTION 2 A simpler solution is possible if we think of using an angle as
a variable. Let O be the angle shown in the bottom figure. Then the area of
the rectangle is
A(0) = (2r cos(0))(r sin(0)) = r²(2sin(0)cos(0)) = r²sin(20)
We know that sin(20) has a maximum value of 1 and it occurs when 20 =
π/2. So A(0) has a maximum value of ² and it occurs when = π/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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