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Name: The figure shows a rectangle inscribed in a right-angled triangle with
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Description: The figure shows a rectangle inscribed in a right-angled triangle with dimensions as shown in the figure.Find x so that the area of the rectangle is as large as possible. 30°

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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The figure shows a rectangle inscribed in a right-angled triangle with dimensions as shown in the figure.
Find x so that the area of the rectangle is as large as possible.

Expert Solution

Step 1

Let y be the length of the opposite side of the rectangle. That is,

Now in triangle ABC,

$\tan 30 ° = \frac{A B}{B C} \Rightarrow \frac{1}{\sqrt{3}} = \frac{y}{1} \Rightarrow y = \frac{1}{\sqrt{3}}$

Step 2

Now let the length of BD be $w$ . Then,

$A D = A B - B D \Rightarrow A D = y - w \Rightarrow A D = \frac{1}{\sqrt{3}} - w$

Now since the triangles ABC and ADE are similar, therefore

$\frac{A D}{D E} = \frac{A B}{B C} \Rightarrow \frac{\frac{1}{\sqrt{3}} - w}{x} = \frac{y}{1} \Rightarrow x y = \frac{1}{\sqrt{3}} - w \Rightarrow x (\frac{1}{\sqrt{3}}) = \frac{1}{\sqrt{3}} - w \Rightarrow x = 1 - \sqrt{3} w \Rightarrow \sqrt{3} w = = 1 - x \Rightarrow w = \frac{1}{\sqrt{3}} (1 - x)$

Now the area of the rectangle is,

$A = x \cdot w [∵ Area of the rectangle = length \times width] \Rightarrow A = x [\frac{1}{\sqrt{3}} (1 - x)] \Rightarrow A = \frac{1}{\sqrt{3}} (x - x^{2})$

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