Question

Let x be the width and y be the height of a norman window. Thus, the semi-circle has radius x/2. We must maximize the area of the window, A = xy + pi/2(x/2)^2. The perimeter of the window is 20 = 2y + x + pi(x/2), and so y = __________________ - x/2 - (pi x)/__________________________.

Help filling the blanks? I assume I have to solve it a certain way, but I have no idea. Thanks.

Answer 1

Find the first and second derivative to find the maximum area.
\[\huge A = xy + \frac{\pi}{2}(\frac{x^2}{4})\]
\[\huge A=xy +\frac{\pi}{8}x^2\]
\[\large y=\frac{20-x-\frac{\pi}{2}x}{2}\]
Therefore:
\[\huge A=\frac{x[20-x-\frac{\pi}{2}x]}{2}+\frac{\pi}{8}x^2\]
\[\huge =\frac{20x-x^2-\frac{\pi}{2}x^2}{2}+\frac{\pi}{8}x^2\]
\[\huge =\frac{80x-4x^2-2\pi x^2+\pi x^2}{8}\]
\[\huge =\frac{80x-4x^2-\pi x^2}{8}\]
\[\large \frac{dA}{dx}=\frac{40-4x-\pi x}{4}\]
\[\large \frac{d^2A}{dx^2}=\frac{-4-\pi}{4}\]
\[\large =-\frac{4+\pi}{4}\]
At S.P's dA/dx=0
\[40-4x-\pi x=0\]
\[40=4x+\pi x\]
\[x(4+\pi)=40\]
\[x=\frac{40}{4+\pi}\]

Answer 2

Sub in x into the equation
\[A=\frac{80x-4x^2-\pi x^2}{8}\]
To find the maximum Area.

Answer 3

gotcha thanks! But I don't know what they want me to put in the blanks!

Answer 4

That's the most simple thing I've ever seen. Never seen someone with this calibre of maths learning this method.
Since you have the given equation of the perimeter
\[\huge 20 = 2y + x + \pi(\frac{x}{2})\]
Rearrange the equation to make y the subject.
\[\huge 2y=20-x-\pi(\frac{x}{2})\]
\[\huge y=\frac{20-x-\frac{\pi x}{2}}{2}\]
Separate the numerator:
\[\huge y=\frac{20}{2}-\frac{x}{2}-\frac{\frac{\pi x}{2}}{2}\]
SImplify:
\[\huge y=10-\frac{x}{2}-[\frac{\pi x}{2}\times \frac{1}{2}]\]
\[\huge =10-\frac{x}{2}-\frac{\pi x}{4}\]

Answer 5

Thank you!