0
$\begingroup$

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.

enter image description here

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?

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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...

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.