Give the x-value and y-value in radical expression. By the way, it says radius before a. Find the dimensions of the rectangle of greatest area that can be inscribed in a circle ofradjus a.

Answered: Find the dimensions of the rectangle of…

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

Chapter8: Areas Of Polygons And Circles

Section8.CR: Review Exercises

Problem 38CR: Prove that if semicircles are constructed on each of the sides of a right triangle, then the area of...

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Give the x-value and y-value in radical expression.

By the way, it says radius before a.

Find the dimensions of the rectangle of greatest area that can be inscribed in a circle of
radjus a.

Expert Solution

Step 1

Let x and y be the length and width of the rectangle inscribe in a circle of radius a.

Then the diagonal of the rectangle is the diameter of the circle,

That is

$x^{2} + y^{2} = {(2 a)}^{2} y = \sqrt{4 a^{2} - x^{2}}$

Step 2

Area of the rectangle is

$A = x y A = x \sqrt{4 a^{2} - x^{2}} \frac{d A}{d x} = - \frac{x^{2}}{\sqrt{4 a^{2} - x^{2}}} + \sqrt{4 a^{2} - x^{2}}$

Now,

$\frac{d A}{d x} = 0 - \frac{x^{2}}{\sqrt{4 a^{2} - x^{2}}} + \sqrt{4 a^{2} - x^{2}} = 0 x^{2} = 4 a^{2} - x^{2} 2 x^{2} = 4 a^{2} x = a \sqrt{2}$

Thus, area of the rectangle is maximum for $x = a \sqrt{2}$

Value of y is

$y = \sqrt{4 a^{2} - (a {\sqrt{2)}}^{2}} = \sqrt{2 a^{2}} y = a \sqrt{2}$

Thus, dimensions of rectangle of greatest area that is inscribed in a circle of radius a are $a \sqrt{2}$ by $a \sqrt{2}$ .

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