Chapter 3.4, Problem 16AYU

Norman Windows A Norman window has the shape of a rectangle surmounted by a semicircle of diameter equal to the width of the rectangle. See the picture. If the perimeter of the window is 20 feet, what dimensions will admit the most light (maximize the area)?

[Hint: Circumference of a circle $=2\pi r;$ area of a circle $=\pi r^2$. where $r$ is the radius of the circle.]

To determine

To calculate: The maximum area of the Norman window with perimeter 20 feet so that maximum light passes through it.

Answer to Problem 16AYU

Solution:

The radius is $r=2.8005$ and the dimensions are x=2.80048 and $y=5.601$.

Explanation of Solution

Given:

The perimeter of the window is 20 feet.

The diagram of the Norman window is

Formula used:

Area of the Norman window is area of the rectangle ply the area of the semicircle.

A=xy+\frac{\pi r^2}{2}, where r is the radius.

Perimeter of the window is $P=2x+y+\pi r$, where $r$ is the radius.

Calculation:

In the given situation, we have $r=\frac{y}{2}$.

Thus, the perimeter becomes

20=2x+y+\pi \left(\frac{y}{2}\right)

\Rightarrow x=10-\frac{y}{2}-\frac{\pi }{4}y

Thus, the area of the window becomes

$A=\left(10-\frac{y}{2}-\frac{\pi }{4}y\right)y+\frac{\pi }{2}{\left(\frac{y}{2}\right)}^2$

$=10y-\frac{y^2}{2}-\frac{\pi }{4}{y}^2+\frac{\pi }{8}{y}^2$

$=10y-\frac{y^2}{2}-\frac{\pi }{8}{y}^2$

Thus, the area is a quadratic equation with $a=-\left(\frac{1}{2}+\frac{\pi }{8}\right),b=10,c=0$

Here, since a is negative, the vertex is the maximum of the quadratic equation.

Thus, the maximum value is present at

$y=-\frac{b}{2a}=-\frac{10}{2\left(-\left(\frac{1}{2}+\frac{\pi }{8}\right)\right)}$

$=\frac{5}{\left(\frac{4+\pi }{8}\right)}=\frac{40}{4+\pi }=5.601$

Then, the maximum value of the area is

A\left(-\frac{b}{2a}\right)=A(5.601)=10(5.601)-\frac{{(5.601)}^2}{2}-\frac{\pi }{8}(5.601{)}^2

=28.005

Thus, the maximum area is $A=28.005feet$.

Here, we have $y=5.601$, now, we need to find the values of x and r.

Therefore, we have

$r=\frac{y}{2}=\frac{5.601}{2}=2.8005$

And x=10-\frac{5.601}{2}-\frac{\pi }{4}(5.601)=2.80048.

Thus, we get the radius, $r=2.8005$ and the dimension x=2.80048 and $y=5.601$.