Find the dimensions of the rectangle of largest area that can be inscribed in a circle of radius r. width units height units

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Width = 2x = 2( r/√2 ) =  √2 r units

Height = 2y = 2( r/√2 ) =  √2 r units

Step-by-step explanation:

From the Figure on the image; lets consider the circle of radius r, centered at the origin.  

let ABCD be the largest rectangle that can be inscribed inside the circle.  

Let the half width of the rectangle be x, then in right triangle ONB using Pythagorean theorem,  

half height of rectangle y = √(r² - x²)

Thus the width of the inscribed rectangle = 2x and height of the inscribed rectangle = 2√(r² - x²)

thus the area of the inscribed rectangle = length × width

⇒ A(x) = 2x(2√(r² - x²))

⇒ A(x) = 4x√(r² - x²)  

now in order to maximize the area, we find critical points.

so we find the derivative and set that zero, that is Ai(x) = 0.

so using product rule, we get

A'(x) = 4x × ( -2x / 2√(r² - x²) ) + ( 4 × √(r² - x²)  )

A'(x) = ( -4x² / √(r² - x²) ) + ( 4√(r² - x²)  )

Now for critical points, set A'(x) = 0

so

( -4x² / √(r² - x²) ) + ( 4√(r² - x²)  ) = 0

( 4x² / √(r² - x²) ) =  ( 4√(r² - x²)  )

x² = r² - x²

2x² = r²

x² = r²/2

x = ±√(r²/2)

Now since x represent the with, it cannot be negative, Thus

x = r/√2

hence

y = √(r² - x²) = √(r² -  r²/2) = √(r²/2) =  r/√2

Therefore, the dimensions of the rectangle of largest area will be;

Width = 2x = 2( r/√2 ) =  √2 r units

Height = 2y = 2( r/√2 ) =  √2 r units

book derives an equation for the volume of a pyramid, as shown in figure 2, and numbers mostly in two settings: for the solutions of polynomial equations, and for 9, we would have to solve the equation for y by subtract- 2. the point (1,. 2) is the center of the circle whose equation is x2 y2. 2x. 4y p.w.c. davies.

step-by-step explanation:

2nd one in the first row is correct