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.

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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?

  • $\begingroup$ I asked, how do I solve to get w in terms of h, or the other way round? Then it would be much easier to solve this problem. Why did someone just vote this question down. If it really is a bad question, I will delete it. $\endgroup$ – cuabanana Apr 20 '13 at 20:47
  • $\begingroup$ For one thing, where in the posted question does the number 300 come from? $\endgroup$ – DJohnM Apr 20 '13 at 20:51
  • $\begingroup$ It seems that png-format image files are not supported here. $\endgroup$ – TonyK Apr 20 '13 at 20:56
  • $\begingroup$ How do I change it to another format? Anyway, if you want to see the image, manually paste the link into a browser tab. $\endgroup$ – cuabanana Apr 20 '13 at 20:57

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 $$h=\frac{300- \frac{600}{4+\pi}\, (1+\frac 12\pi)}{2}\approx 42$$

  • $\begingroup$ I tried. Didn't work. $\endgroup$ – cuabanana Apr 21 '13 at 1:32

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