 Khaleesi Herbert

2021-08-06

A Norman window of maximum area can be constructed by adjoining a semicircle to the top of an ordinary rectangular window. The dimensions of the Norman window would be such that the radius is x/2 and the total perimeter is 22 feet. funblogC

Let represent the measure of the vertical dimension of the rectangular portion of the window. Let represent the horizontal dimension of the window. is also the diameter of the semi-circular part of the window. The perimeter of the window is then: $22=x+2y+\frac{1}{2}\pi x$
${P}_{Total}=\left({P}_{Rectangle}-x\right)+\frac{{P}_{Circle}}{2}$
$10-\frac{1}{2}x-\frac{1}{5}\pi x=y$
$A=xy+\frac{\pi {\left(\frac{1}{2}x\right)}^{2}}{2}$
${A}_{Total}={A}_{Retangle}+\frac{{A}_{Circle}}{2}$
$A=x\left(10-\frac{1}{2}x-\frac{1}{5}\pi x\right)+\frac{\pi {\left(\frac{1}{2}x\right)}^{2}}{2}$
Plug in $10-\frac{1}{2}x-\frac{1}{5}\pi x$ for $y$
$A=10x-\frac{1}{2}{x}^{2}-\frac{\pi {x}^{2}}{10}$
$\frac{dA}{dX}=10-x-\frac{\pi x}{5}$
Power rule $\frac{d}{dx}{x}^{n}=n{x}^{n-1}$
$0=10-x-\frac{\pi x}{5}$
Set $\frac{dA}{dx}$ equal to 0
$x+\frac{\pi x}{5}=10$
Add $x+\frac{\pi x}{5}$ to get x alone
$x\left(1+\frac{\pi }{5}\right)=10$
$x=\frac{10}{1+\frac{\pi }{5}}$
$x=\frac{40}{4+\pi }$
$\frac{{d}^{2}A}{{dx}^{2}}=-\frac{\left(4+\pi \right)}{4}<0$ Solve for y:
$y=10-\frac{\left(2+\pi \right)x}{4}$
Hence $x=\frac{40}{4+\pi }$ is the x-coordinate of the maximum point. The y dimensoin and the semi-circle redius are calculated as above.

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