Anorman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. what is the area of the largest possible norman window with a perimeter of 25 feet

43.75 ft²

Step-by-step explanation:

     Let r = the radius of the semicircle

   and h = the height of the rectangle

Then 2r = the width of the window

The formula for the perimeter of a circle is C = 2πr,

so, πr = the perimeter of the semicircle

The perimeter of the window is

P = πr + 2h + 2r = 25

      2h + (π +2)r = 25

                       h = ½[25 - (π + 2)r]

(1)                   h = 12.5 - (π/2 +1)r

The formula for the area of a circle is A= πr²,  so

½πr² = the perimeter of the semicircle

The area of the window is

(2) A = ½πr² + 2rh

Substitute (1) into (2).

    A = ½πr² + 2r[12.5 - (π/2 +1)r] = ½πr² + 25r - (π +2)r²

    A = 25r - (π + 2 - π/2)r²

(3) A =  -(π/2 + 2)r² + 25r

This is the equation for a downward opening parabola.

One way to find the vertex is to set the first derivative equal to zero.

dA/dr = -2(π/2 + 2)r + 25 = 0

               -(π    + 4)r  + 25 = 0

               -(π     + 4)r        = -25

                                     r = 25/(π + 4)

(4)                                 r ≈ 3.50 ft

The maximum area occurs when r = 3.50 ft.  

Substitute (4) into (1).

h = 12.5 - (π/2 +1)(3.50) = 12.5 - (2.571× 3.50) = 12.5 - 9.00 = 3.50  

(4) h = 3.50 ft

Substitute (4) into (2)

A =  1.571(3.50)² + 2×3.50×3.50 = 19.25 + 24.50

A = 43.75 ft²

The area of the largest possible Norman window with a perimeter of 25 ft is 43.75 ft².

it would be 6xy and -16xy.

so what is the problem what are they asking you to solve

step-by-step explanation:

so what is the problem what are they asking you to solve