Question

A Norman window is a rectangle with a semicircle on top. Suppose that the perimeter of a particular Norman window is to be 56 ft. What should its dimensions be in order to allow the maximum amount of light to enter through the​ window?

Answer 1

Answer:

Dimensions of the window in order to allow maximum light is $x=\dfrac{56}{4+\pi}$ and $y=\dfrac{56}{4+\pi}$

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,  

$Perimerter = AB + BC + CD + arc\:AD$

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 $C=2\pi r$. So half of it will be  

$C=\dfrac{2\pi r}{2}$

$C=\dfrac{2\pi x}{2}$

$C=\pi x$

So, $arc\:AD = \pi x$

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  

$\therefore BC=2x,AB=CD=y$  

Substituting the value,  

Perimeter = AB + BC + CD + arc AD  

$56=y+2x+y+\pi x$  

$56=2y+2x+\pi x$  

To calculate value of y,subtracting 2x and $\pi x$ on both sides,  

$56-2x-\pi x=2y$  

Dividing by 2,

$28-x-\dfrac{\pi x}{2}=y$  

Now calculate area of window.  

Area = Area of rectangle + Area of semicircle

From diagram,

$Area = width\times length + \dfrac{1}{2}\times\pi r^{2}$

$Area = 2x\times y + \dfrac{1}{2}\times\pi x^{2}$

$Area = 2xy + \dfrac{\pi x^{2}}{2}$

Substituting value of y in above equation,

$Area = 2x\left (28-x-\dfrac{\pi x}{2} \right ) + \dfrac{\pi x^{2}}{2}$

Simplifying,  

$Area = 56x-2x^{2}-\pi x^{2}+ \dfrac{\pi x^{2}}{2}$

$Area = 56x-2x^{2}- \dfrac{\pi x^{2}}{2}$

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,  

$\dfrac{dA}{dx}=\dfrac{d}{dx}\left(56x\right)-\dfrac{d}{dx}\left ( 2x^{2} \right )-\dfrac{d}{dx}\left ( \dfrac{\pi x^{2}}{2} \right )$

Applying constant multiple rule of derivative,  

$\dfrac{dA}{dx}=56\dfrac{d}{dx}\left(x\right)-2\dfrac{d}{dx}\left ( x^{2} \right )-\dfrac{\pi}{2}\dfrac{d}{dx}\left (x^{2}\right )$

Applying power rule of derivative,

$\dfrac{dA}{dx}=56\left(1x^{1-1}\right)-2\left(2x^{2-1}\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x^{2-1}\right )$

$\dfrac{dA}{dx}=56\left(1\right)-2\left(2x\right)- \dfrac{\pi}{2}\dfrac{d}{dx}\left (2x\right )$

$\dfrac{dA}{dx}=56-4x-\pi x$

Now find the critical number by solving as follows,

$\dfrac{dA}{dx}=0$

$56-\left(4+\pi\right)x =0$

$56=\left(4+\pi \right)x$

$\dfrac{56}{4+\pi} =x$

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.  

$Area = 56\left (\dfrac{56}{4+\pi} \right )-2\left (\dfrac{56}{4+\pi} \right )^{2}- \dfrac{\pi \left (\dfrac{56}{4+\pi} \right )^{2}}{2}$

Calculating the above expression,  

$Area = \dfrac{1568}{4+\pi }$

So area is greater than 0.  

Now calculate value of y,  

$28-\dfrac{56}{4+\pi}-\dfrac{\pi}{2}\left ( \dfrac{56}{4+\pi} \right )=y$

$\dfrac{56}{4+\pi}=y$

Hence dimensions are $x=\dfrac{56}{4+\pi}$ and $y=\dfrac{56}{4+\pi}$