Question

A norman window has the shape of a rectangle surmounted by a semicircle. (thus the diameter of the semicircle is equal to the width of the rectangle.) if the perimeter of the window is 16 ft, find the value of x so that the greatest possible amount of light is admitted. (give your answer correct to two decimal places.) x = ft

Answer 1

Answer-

For x = 2.24, greatest possible amount of light will be admitted.

Solution-

According to the question, the perimeter of the window is 16 ft

From the attachment,

Perimeter of the window = Perimeter of the rectangle-Width of the rectangle + Curved perimeter of the semi-circle

So,

$\Rightarrow (2x+y+y+2x)-(2x)+(\pi x)=16\\\\\Rightarrow 2x+2y+\pi x=16\\\\\Rightarrow 2y=16-2x-\pi x\\\\\Rightarrow y=\dfrac{16-2x-\pi x}{2}\\\\\Rightarrow y=8-x-\dfrac{\pi x}{2}$

As we have to find the value of x so that the greatest possible amount of light is admitted, so we have to calculate the maximum area for which the perimeter is 16 ft.

$\text{Area of the window}=\text{Area of the rectangle + Area of the semi-circle}$

So,

$A=(2x\times y)+(\dfrac{\pi x^2}{2})$

Putting the value of y in terms of x,

$\Rightarrow A=2x(8-x-\dfrac{\pi x}{2})+\dfrac{\pi x^2}{2}$

$\Rightarrow A=16x-2x^2-\pi x^2+\dfrac{\pi x^2}{2}$

$\Rightarrow A=16x-2x^2-\dfrac{\pi x^2}{2}$

$\Rightarrow A'=16-4x-\pi x$

Now, equating the first derivative of A to 0,

$\Rightarrow A'=0$

$\Rightarrow 16-4x-\pi x=0$

$\Rightarrow 16-x(4+\pi)=0$

$\Rightarrow x(4+\pi)=16$

$\Rightarrow x=\dfrac{16}{4+\pi}$

$\Rightarrow x\approx 2.24$

Therefore, for x= 2.24, greatest possible amount of light will be admitted.

Answer 2

If the perimeter of the window is 16 ft, the value of x is 8,4 ft * 4,2 ft, so that the greatest possible amount of light is admitted.

Explanation:

Semicircle is a one-dimensional locus of points that forms half of a circle. The full arc of a semicircle always measured as 180°. Whereas a rectangle is a quadrilateral with four right angles.

The greatest possible amount of light will be admitted if the are of the window is maximum.

Let x denote as half the width of rectangle, therefore x is the radius of semicircle and let y as the height of the rectangle.

Window perimeter

$2x+2y+\frac{1}{2} (2x\pi) = 2y + (2+\pi)x$

Therefore  

$30 = 2y+(2+\pi)x$

$y=\frac{30-(2+\pi)x}{2} =15-(1+\frac{\pi}{2})x$

Window area is

A = area of rectangle + area of semicircle

$A = (2x)y+\frac{1}{2} (\pi x^2)$

Then by substitute the value of y we get

$A(x)=2x[15-(1+\frac{\pi}{2})x] + \frac{\pi x^2}{2} \\A(x)=30x-2x(1+\frac{\pi}{2})x + \frac{\pi x^2}{2} \\A(x)=30x-2x^2-\pi x^2+ \frac{\pi x^2}{2} \\A(x)=30x-(2+\frac{\pi }{2})x^2$

We need maximizing A(x) to find x value, for which it is maximum. Therefore we will equate the A'(x) is equal to zero.

$A'(x) = 30-2(2+\frac{\pi }{2} ) x = 30-(4+ \pi)x\\A'(x)=0 \\30-(4+ \pi) x = 0\\30 = (4+ \pi)x\\x = \frac{30}{4+ \pi} = 4.2$

The condition of domain are x>0, y>0, and A(x)>0

Therefore the domain is $x E (0,\frac{15}{1+\frac{\pi}{2} } )$

We will find the value of A at x=0, $x = \frac{15}{1+\frac{\pi}{2} }$ and x = \frac{30}{4+\pi}

We find that A(0) = 0

$A(\frac{30}{4+\pi} ) = A(4,2) = 63\\A(\frac{15}{1+\frac{\pi}{2} } ) =53.5$

Window area is maximum when $x = \frac{30}{4+ \pi}$

When $x = \frac{30}{4+ \pi}$, by using $Eqn(1)y = 15-(1+\frac{\pi}{2} )*(\frac{30}{4+\pi} ) = \frac{30}{4+\pi}$

Therefore maximum light is admitted through the window.

When radius of semi-circular part is $\frac{30}{4+\pi} = 4.2 ft$

Therefore the dimensions of the rectangular part of the window are 8,4 ft * 4,2 ft

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