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.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Solution 1:**
Let's take the semicircle to be the upper half of the circle \( x^2 + y^2 = r^2 \) with center at the origin. The rectangle is inscribed such that it has two vertices on the semicircle and two vertices on the \( x \)-axis.
Let \( (x, y) \) be the vertex that lies in the first quadrant. The rectangle then has sides of lengths \( 2x \) and \( y \), and its area is \( A = 2xy \).
To eliminate \( y \), use the fact that \( (x, y) \) lies on the circle \( x^2 + y^2 = r^2 \) and so \( y = \sqrt{r^2-x^2} \). Thus,
\[ A = 2x\sqrt{r^2-x^2}. \]
The domain of this function is \( 0 \leq x \leq r \). Its derivative is
\[
A' = 2\sqrt{r^2-x^2} - \frac{2x^2}{\sqrt{r^2-x^2}} = \frac{2(r^2-2x^2)}{\sqrt{r^2-x^2}}
\]
which is 0 when \( 2x^2 = r^2 \), that is, \( x = \frac{r}{\sqrt{2}} \) (since \( x \geq 0 \)).
Therefore, the area of the largest inscribed rectangle is
\[ A\left(\frac{r}{\sqrt{2}}\right) = 2\frac{r}{\sqrt{2}} \times \frac{r}{\sqrt{2}} = r^2. \]
**Solution 2:**
A simpler solution uses an angle as a variable. Let \( \theta \) be the angle shown in the figure. Then the area of the rectangle is
\[ A(\theta) = (2r\cos(\theta))(r\sin(\theta)) = r^2(2\sin(\theta)\cos(\theta)) = r^2\sin(2\theta). \]
The maximum value of \( \sin(2\theta) \) is 1, occurring when \( 2\theta = \frac{\pi}{2}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5abad2ef-7d64-4544-b6f7-dd3dfbff74f8%2F2de87cda-18b4-4285-8a54-050f0897650a%2F4hbgc8_processed.png&w=3840&q=75)
Transcribed Image Text:**Solution 1:**
Let's take the semicircle to be the upper half of the circle \( x^2 + y^2 = r^2 \) with center at the origin. The rectangle is inscribed such that it has two vertices on the semicircle and two vertices on the \( x \)-axis.
Let \( (x, y) \) be the vertex that lies in the first quadrant. The rectangle then has sides of lengths \( 2x \) and \( y \), and its area is \( A = 2xy \).
To eliminate \( y \), use the fact that \( (x, y) \) lies on the circle \( x^2 + y^2 = r^2 \) and so \( y = \sqrt{r^2-x^2} \). Thus,
\[ A = 2x\sqrt{r^2-x^2}. \]
The domain of this function is \( 0 \leq x \leq r \). Its derivative is
\[
A' = 2\sqrt{r^2-x^2} - \frac{2x^2}{\sqrt{r^2-x^2}} = \frac{2(r^2-2x^2)}{\sqrt{r^2-x^2}}
\]
which is 0 when \( 2x^2 = r^2 \), that is, \( x = \frac{r}{\sqrt{2}} \) (since \( x \geq 0 \)).
Therefore, the area of the largest inscribed rectangle is
\[ A\left(\frac{r}{\sqrt{2}}\right) = 2\frac{r}{\sqrt{2}} \times \frac{r}{\sqrt{2}} = r^2. \]
**Solution 2:**
A simpler solution uses an angle as a variable. Let \( \theta \) be the angle shown in the figure. Then the area of the rectangle is
\[ A(\theta) = (2r\cos(\theta))(r\sin(\theta)) = r^2(2\sin(\theta)\cos(\theta)) = r^2\sin(2\theta). \]
The maximum value of \( \sin(2\theta) \) is 1, occurring when \( 2\theta = \frac{\pi}{2}
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