A Norman window has the shape of a rectangle with a semi circle on top; diameter of the semicircle exactly matches the width of the rectangle. Find the dimensions of the Norman window whose perimeter is 300 in that has maximal area.
The area of the semicircle would be $(\pi(w/2)^2)/2$. The area of the rectangle would be $hw$. I know that the perimeter is 300 in, and that the perimeter would be $2h+w+(w\pi) = 300$. How would I write $h$ in terms of $w$, and then solve for $h$ to specify the dimensions? The total area would be the 2 sub-areas added together. I would have to take the derivative of the combined areas to solve for the width and height. What would be the proper steps for doing this?
2 Answers
OK, I spotted the error: when you wrote the expression (which Anil Baseski repeated) for the perimeter, you used the circumference of a circle written as $C = \pi d$. However, the "lunette" of the Norman window is only a semi-circle, so the perimeter equation should be $p = 2h + w + \frac{\pi}{2} w = 300$ . The corrections are then
$$h = 150 - \left(\frac{\pi + 2}{4}\right)w , A = 150w - \left(\frac{4 + \pi}{8}\right)w^2,$$
$$\frac{dA}{dw} = 150 - \left(\frac{4 + \pi}{4}\right)w = 0 \Rightarrow w = \frac{600}{4 + \pi} \approx 84.0$$
$$ \Rightarrow h \approx 150 - \left(\frac{\pi + 2}{4}\right) \cdot 84.0 \approx 42.0.$$
Needless to say, such windows in actual use are designed for esthetics and not maximal area: this window is way too wide, relative to its total height, to be appealing...
Area of the window $$A=\frac{\pi }{8}w^2+wh$$ and the perimeter $$P=2\, h+w+\frac12\pi\, w$$ The problem is $$max_{w,h}\ A\qquad s.t.\, P=300$$ Using perimeter constraint you can eliminate for h $$h=\frac{300- w\, (1+\frac12\pi)}{2}$$ The area is then $$A=\frac{\pi }{8}w^2+150\,w-\frac{1+\frac12\pi}{2}w^2=-\frac{4+\pi}{8}w^2+150\,w$$ By taking the derivative and setting to zero $$\frac{dA}{dw}=-\frac{4+\pi}{4}w+150=0\Rightarrow w=\frac{600}{4+\pi}\approx 84$$ and $$h=\frac{300- \frac{600}{4+\pi}\, (1+\frac 12\pi)}{2}\approx 42$$
