What Is the Optimal Design for a Norman Window to Maximize Light?

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Homework Help Overview

The discussion revolves around optimizing the design of a Norman window, which consists of a rectangle topped by a semicircle, with a fixed perimeter of 30 ft. Participants are tasked with modeling the area of the window and determining the dimensions that maximize light admission.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive an area function based on given dimensions and perimeter constraints, expressing uncertainty about maximizing the area without using calculus. Other participants question the nature of the polynomial formed and its graphical properties, exploring the implications of its shape on the number of maxima.

Discussion Status

Participants are actively engaging with the problem, with some providing insights into the characteristics of the polynomial and its graphical representation. The original poster has made progress in calculating potential dimensions but expresses confusion regarding the methodology presented in their textbook.

Contextual Notes

The original poster notes a restriction on using derivatives or calculus, which influences their approach to finding the maximum area. There is also mention of discrepancies in the area equation, leading to further exploration of its implications.

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Homework Statement



a norman window has the shape of a rectangle surmounted by a semicircle. A norman window with the perimeter 30 ft. is to be constructed.

a.) find a function that models the area of the window.
b.) find the dimensions of the window that admit the greatest amount of light

Homework Equations


part a.) area= LW (this problem, it will be xy, with x as the width)
semicircle circumference= 1/2\Pix s
part b.) i don't know and I cannot use derivatives or calculus, which is why I am having
trouble here.

The Attempt at a Solution



a.)
P= x + 2y + 1/2\Pix = 30
2y = 30 - 1/2\Pix - x
y = 15 - 1/2x - 1/4\Pix

A= (x)(15 - 1/2x - 1/4\Pix) + 1/2\Pi(1/2x)2
(x)(15 - 1/2x - 1/4\Pix) + 1/8\Pix2
hence A= 15x - 1/2x2 - 1/8\Pi2

b.)
i know the area equation is correct. I have no idea as to how to begin to figure out the max dimensions for the light. i would appreciate any clues. and sorry if the pi isn't looking right.
 
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Assuming your area equation is correct,

This type polynomial has a special name; what is it?

What do we call the graph of this type of polynomial?

What do we know about the number of (in this case) maximums this graph has, and how do we find it/them?
 
well, on the area equation, i did write it wrong, it was supposed to be:

A = 15x - 1/2x2 - 1/8\Pix2

anyway, there are 2 x2, so i don't know. i can only see there being one maximum. i don't know, maybe its because the pi is in there and there are 2 numbers of the same degree, i just don't know because i have never done a problem like this. i put the area equation in the calculator, like i should have done in the beginning, with the pi and everything and the max is x=8.40. i know that is also correct and i don't know why i didnt try that before.

so, putting it into the y (or height) formula, 15 - 1/2(8.40) - 1/4\Pi(8.40) = 4.20

and the max area would be 465.51ft2

my book gives a long, completely different way of going about the solution that i can't figure out by looking at it. I am not sure if any of this is what you meant, but thanks for taking the time to reply and get me thinking.
 
The last two terms in your equation for A can be written

- (1/2 + (1/8)pi) x2

The graph of your equation is a parabola which opens down (the coefficient of x2 is negative) and thus has one maximum.

You were on the right track! This is not an easy problem.
 

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