Respuesta :
The exact value of x (in ft) so that the greatest possible amount of light is admitted is 8.96 ft
According to the question,
A norman window has the shape of a rectangle surmounted by a semicircle.
Let The width of rectangle be x and height be y
It is given that the diameter of the semicircle is equal to the width of the rectangle
Therefore , Diameter of circle = x
Radius = x/2
Perimeter of window = 32 = Perimeter of rectangle + Perimeter of semi-circle
=> 32 = Width + 2height + πr
=> x + 2y + πx/2 = 32
=> 2x + 4y + πx = 64
Solving for y,
=> 4y = 64 - 2x - πx
=> y = 64 - 2x - πx/ 4 ----------(1)
Area of the window = Area of rectangle + area of semi-circle
=> A = Width × hiegth + πx²/8
=> A = xy + πx²/8
Substituting the value of y from equation (1)
=> A = x ( 64 - 2x - πx/ 4 ) + πx²/8
[tex]A = x ( \frac{64 - 2x -\pi x}{ 4} ) + \frac{\pi x^2}{8}\\A = ( \frac{ 64x - x^2 ( 2 + \pi)}{4 } ) + \frac{\pi x^2}{8}[/tex]
Differentiating A w.r.t x
=> [tex]\frac{dA}{dx} = (\frac{64 - 2x(2 + \pi)}{4} ) + \frac{\pi x}{4}[/tex]
Now , To find a maximum area
dA/dx = 0
=> [tex]\frac{dA}{dx} = (\frac{64 - 2x(2 + \pi)}{4} ) + \frac{\pi x}{4} = 0[/tex]
Solving for x ,
64 - 4x - 2π + πx = 0
=> x = 64 / 4 + π ------(2)
Differentiating A again w.r.t x
=> d²A / dx² = -2(2 + π)/4 + π/4
=> -π+4 / 4 which is less than zero
Therefore,
Area is maximum when x = 64 / 4 + π
The exact value of x (in ft) so that the greatest possible amount of light is admitted is 8.96 ft
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