A norman 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

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Answer:

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.

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

What is the maximum area?

If r and h are the radius of the semicircle and height of the rectangle respectively, it means;

2r is the width of the window

Formula for the perimeter of a semi circle is; C = πr

Thus the perimeter of the window is expressed as;

P = πr + 2h + 2r = 25

making h the subject gives;

h = 12.5 - (¹/₂π +1)r   ------(1)

Formula for the area of the semi-circle is; A = ½πr²

The area of the window is;

A = ½πr² + 2rh     -------(2)

Putting the expression of h for h in eq 2 gives;

A = ½πr² + 2r[12.5 - (¹/₂π +1)r]

A = ½πr² + 25r - (π +2)r²

A = 25r - (π + 2 - π/2)r²

A = -(¹/₂π + 2)r² + 25r

Let us find the vertex by finding the first derivative and equating to zero.

dA/dr = -2(¹/₂π + 2)r + 25 = 0

dA/dr = -(π + 4)r  + 25 = 0

dA/dr =  -(π + 4)r  

Since Perimeter = dA/dr, then;

-(π + 4)r  = -25

solving for r gives;

r ≈ 3.50 ft

The maximum area will occur when r = 3.50 ft.  

Putting 3.5 for r in eq 1 gives;

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

h = 12.5 - 9.00

h = 3.50 ft

put r = 3.5 and h = 3.5 in eq 2 to get;

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

A = 43.75 ft²

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