Question

A Norman window has the shape of a rectangle surmounted by a semicircle. Suppose the outer perimeter of such a window must
be 600 cm. In this problem you will find the base length a which will maximize the area of such a window. Use calculus to find
an exact answer.
When the base length is zero, the area of the window will be zero. There is also a limit on how large scan be: when x is large
enough, the rectangular portion of the window shrinks down to zero height. What is the exact largest value of r when this
occurs?

Largest x = ?

Answer 1

The base length that will maximize the area for such a window is 168.03 cm. The exact largest value of x when this occurs is 233.39 cm

Suppose we make an assumption that:

• (x) should be the width of the rectangle base;
• (h) should be the height of the rectangle

Also, provided that the diameter of the semi-circle appears to be the base of the rectangle, then;

• the radius  $\mathbf{r = \dfrac{x}{2}}$  

and, the perimeter of the window can now be expressed as:

$\mathbf{x + 2h + \pi r = x + 2h + \dfrac{\pi x }{2}}$

$\mathbf{= \Big ( 1 + \dfrac{\pi}{2}\Big) x + 2h}$

Given that the perimeter = 600 cm

∴

$\mathbf{ \Big ( 1 + \dfrac{\pi}{2}\Big) x + 2h= 600}$

$\mathbf{ h = 300 - \Big( \dfrac{1}{2} + \dfrac{\pi}{4}\Big) x}$

Since h > 0, then:

$\mathbf{ h = 300 - \Big( \dfrac{1}{2} + \dfrac{\pi}{4}\Big) x>0}$

By rearrangement and using the inverse rule:

$\mathbf{ x< \dfrac{ 300}{\Big( \dfrac{1}{2} + \dfrac{\pi}{4}\Big) } }$

$\mathbf{ x= \dfrac{ 1200}{\Big( 2 +\pi \Big) } }$

$\mathbf{ x= 233.39 \ cm }$

Thus, the largest length x = 233.39 cm

However, the area of the window is given as:

$\mathbf{A(x) = xh + \dfrac{1}{2} \pi r^2}$

$\mathbf{A = x \Big [ 300 - \Big ( \dfrac{1}{2}+\dfrac{1}{4} \Big) x \Big ] +\dfrac{1}{2}\pi \Big(\dfrac{x}{2} \Big )^2}$

$\mathbf{A (x) = 300x - \Big( \dfrac{1}{2} + \dfrac{\pi}{8}\Big) x^2 \ cm^2}$

Now, at maximum, when the area A = 0. Taking the differentiation, we have:

$\mathbf{\dfrac{d}{dx} 300x - \dfrac{d}{dx} \Big( \dfrac{1}{2} + \dfrac{\pi}{8}\Big) x^2 \ =0}$

$\mathbf{ 300 - 2x \Big( \dfrac{1}{2} + \dfrac{\pi}{8}\Big) \ =0}$

Making x the subject of the formula, we have:

$\mathbf{x = \dfrac{1200}{4 +\pi}}$

x = 168.03 cm

Taking the second derivative:

$\mathbf{\dfrac{d}{dx} \Big [300 -2x \Big( \dfrac{1}{2} + \dfrac{\pi}{8}\Big) \Big]}$

$\mathbf{= -2 \Big( \dfrac{1}{2}+\dfrac{\pi}{8}\Big ) <0}$

Therefore, we can conclude that the maximum area that exists for such a window is 168.03 cm

Learn more about derivative here:

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