Answer :
Answer:
Dimensions of the window in order to allow maximum light is [tex] x=\dfrac{56}{4+\pi} [/tex] and [tex] y=\dfrac{56}{4+\pi}[/tex]
Step-by-step explanation:
Consider following Norman window, assuming ABCD as rectangle and arc AD as semicircle with center at E and radius r. (Refer attachment)
Given that perimeter of window is 56 ft. Therefore perimeter of window is given as,
[tex]Perimerter = AB + BC + CD + arc\:AD [/tex]
Calculate arc AD as follows,
Let, x denote radius of semi circle. That is, r=x
Since AD is the diameter of semi circle
So AD = 2 r = 2 x.
Now perimeter of semicircle is equal to circumference of semicircle, so calculate circumference of semicircle as follows
Circumference of circle is [tex]C=2\pi r[/tex]. So half of it will be
[tex]C=\dfrac{2\pi r}{2}[/tex]
[tex]C=\dfrac{2\pi x}{2}[/tex]
[tex]C=\pi x [/tex]
So, [tex]arc\:AD = \pi x[/tex]
Calculate AB , BC and AD as follows,
Consider rectangle ABCD,
Since, AD is the diameter of semi circle which is also one of the side of the rectangle.
So AD = 2 r = 2 x.
Since AD is parallel to BC. Therefore, AD=BC=2x. Also length of rectangle be y
[tex]\therefore BC=2x,AB=CD=y[/tex]
Substituting the value,
Perimeter = AB + BC + CD + arc AD
[tex]56=y+2x+y+\pi x[/tex]
[tex]56=2y+2x+\pi x[/tex]
To calculate value of y,subtracting 2x and [tex]\pi x[/tex] on both sides,
[tex]56-2x-\pi x=2y[/tex]
Dividing by 2,
[tex]28-x-\dfrac{\pi x}{2}=y[/tex]
Now calculate area of window.
Area = Area of rectangle + Area of semicircle
From diagram,
[tex]Area = width\times length + \dfrac{1}{2}\times\pi r^{2}[/tex]
[tex]Area = 2x\times y + \dfrac{1}{2}\times\pi x^{2}[/tex]
[tex]Area = 2xy + \dfrac{\pi x^{2}}{2}[/tex]
Substituting value of y in above equation,
[tex]Area = 2x\left (28-x-\dfrac{\pi x}{2} \right ) + \dfrac{\pi x^{2}}{2}[/tex]
Simplifying,
[tex]Area = 56x-2x^{2}-\pi x^{2}+ \dfrac{\pi x^{2}}{2}[/tex]
[tex]Area = 56x-2x^{2}- \dfrac{\pi x^{2}}{2}[/tex]
In order to find the maximum of area function, differentiate the equation with respect to x and find the critical points.
Applying difference rule of derivative,
[tex]\dfrac{dA}{dx}=\dfrac{d}{dx}\left(56x\right)-\dfrac{d}{dx}\left ( 2x^{2} \right )-\dfrac{d}{dx}\left ( \dfrac{\pi x^{2}}{2} \right )[/tex]
Applying constant multiple rule of derivative,
[tex]\dfrac{dA}{dx}=56\dfrac{d}{dx}\left(x\right)-2\dfrac{d}{dx}\left ( x^{2} \right )-\dfrac{\pi}{2}\dfrac{d}{dx}\left (x^{2}\right ) [/tex]
Applying power rule of derivative,
[tex]\dfrac{dA}{dx}=56\left(1x^{1-1}\right)-2\left(2x^{2-1}\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x^{2-1}\right ) [/tex]
[tex]\dfrac{dA}{dx}=56\left(1\right)-2\left(2x\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x\right ) [/tex]
[tex]\dfrac{dA}{dx}=56-4x-\pi x[/tex]
Now find the critical number by solving as follows,
[tex]\dfrac{dA}{dx}=0[/tex]
[tex]56-\left(4+\pi\right)x =0[/tex]
[tex]56=\left(4+\pi \right)x[/tex]
[tex] \dfrac{56}{4+\pi} =x [/tex]
Since there is only one critical point, directly substitute the value of x into equation of A. If value of A is greater than 0, then the area is maximum at critical point.
[tex]Area = 56\left (\dfrac{56}{4+\pi} \right )-2\left (\dfrac{56}{4+\pi} \right )^{2}- \dfrac{\pi \left (\dfrac{56}{4+\pi} \right )^{2}}{2}[/tex]
Calculating the above expression,
[tex] Area = \dfrac{1568}{4+\pi }[/tex]
So area is greater than 0.
Now calculate value of y,
[tex] 28-\dfrac{56}{4+\pi}-\dfrac{\pi}{2}\left ( \dfrac{56}{4+\pi} \right )=y [/tex]
[tex] \dfrac{56}{4+\pi}=y [/tex]
Hence dimensions are [tex] x=\dfrac{56}{4+\pi} [/tex] and [tex] y=\dfrac{56}{4+\pi}[/tex]
