Area of Norman Window: Function of Width x

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Homework Help Overview

The problem involves a Norman window, which consists of a rectangle topped by a semicircle. The task is to express the area of the window as a function of its width, given that the perimeter is 30 ft.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the correct formulation of the perimeter, questioning the initial assumption that it is simply the sum of the rectangle's sides and width.
  • There are attempts to express the area in terms of width and length, with some participants suggesting the need to incorporate the semicircular part into the perimeter calculation.
  • Several participants express confusion about how to relate the dimensions of the rectangle and semicircle to the given perimeter constraint.
  • Questions arise regarding the correct expressions for the area of both the rectangle and the semicircle, as well as how to consolidate variables into a single function.

Discussion Status

The discussion is ongoing, with various interpretations and approaches being explored. Some participants have proposed equations for the area and perimeter, while others are clarifying the relationships between the dimensions. There is no explicit consensus yet, but several productive lines of reasoning are being developed.

Contextual Notes

Participants are working under the constraint that the total perimeter of the window is 30 ft, which influences their calculations and formulations. There is also a focus on expressing the area in terms of a single variable, which adds complexity to the discussion.

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Homework Statement




A norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 ft, express the area A of the window as a function of the width x of the window.

Homework Equations



2l +2w=30

The Attempt at a Solution


If this were simply a rectangle,

I can solve for l in terms of w.

l=15-w

A=w(15-w)


but I don't know how to handle the semicircle part.

Homework Statement





Homework Equations



(pir^2)/2


The Attempt at a Solution

 
Last edited:
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First, 2l+2w is NOT 30 ft. The perimeter includes the curved (semi-circle) part, so the perimeter is the sum of the lengths of the two sides, the base (the width 'x'), and the semi-circle.
 
So how do I convert that into function form?
 
Just add the quantities I stated.

sum of the lengths of the two sides, the base (the width 'x'), and the semi-circle.

Length = l, width = x, perimeter of semi-circle = ?
 
But the perimeter of the entire window is 30 ft. The length and width of the rectangle/semicircle are unknown.
 
OK, here's what I came up with.

A(w)=w(30-w-pir^2)/2
 
starchild75 said:
But the perimeter of the entire window is 30 ft. The length and width of the rectangle/semicircle are unknown.
You are asked to find a function; not to compute a value. A value of width or length is not needed.

r=radius, L=length, p=pi
area= L*2*r + 0.5*p*r^2;
perimeter= 2*r + 2*L + p*r = 30

You only have two variables there. Any found value of r will determine the corresponding value for L (the other dimension of the rectangle).. There is enough information to find the function of area based on ONE variable, the radius, r.
I used 2*r as the width, so you may want to rewrite some of the above shown equations based on "x" instead of 2*r.
 
So in solving for r, I get (30-2l-ps)/2

where ps is perimeter of semicircle.

But I want that in r in terms of l right?
 
putting the equation in terms of width, I got w(30-r-Ps)

P being perimeter of the semicircle.

How does that look?
 
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  • #10
Maybe changing the variables made the information confusing.
Let us try:
x=width of rectange, same as diameter of the semicircle;
L=length of rectangle
p= Pi=close to 3.1415...
A= area

A= 0.5*p*(x/2)^2 + x*L
and
30 = p*(x/2) + 2*L + x

You want A in terms of x. Use the second equation and substitute expression for L into the first equation.
 
  • #11
relate the length or the arc of the semi circle to the variable of the rectangle (w and or l)

the equation of a circle's cicomference is C=2pi*r correct? you soul not need any more information
 

Attachments

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  • #12
Now I am completely lost. The diameter would be c/pi? The perimeter would be pir-2r+2r+2l. or pir+2l?
 
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  • #13
there is a picture attached to my post, use the picture

if you can't see it i will explain it.

i denoted the bottom of the window as W
the sides as L
and the arc perimiter= unknown

however we know the radius of the arc is just W/2
so r=W/2

the formula of the perimiter of a circle is C=2pi*r
so given that R=W/2 what is the circumference?
*hint for next step...this is not a circle it is a semi circle*
what is the cirumference of the semi circle
 
  • #14
attachment pending approval. You are saying use the circumference minus diameter plus diameter (the base) plus the two lengths for perimeter?
 
  • #15
Ok see what you think of this.

A(w)= w((30-w-pir)/2)+(3-w-2H)/pi


w=width or diameter

H=height

r=radius
 
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  • #16
not exactly sure what you are saying. I might have worded my question incorrectly.

i meant the circumference of the semi circle (not including the base...only the arc)
which is essentialy half of the circumference of the circle.
wha did you get fo the circumference of the circle?


starchild75 said:
attachment pending approval. You are saying use the circumference minus diameter plus diameter (the base) plus the two lengths for perimeter?
 
  • #17
i appologize humbly i may have misundertsood the question.

so i read the questionover again and now it has confused me...its a rectangle with a semi circle up top right, and you want to find the area of the circle (as an equation) but your given that 2l+2w=30 (or did you do that?) because the 2L i understand, but shouldn't it be 2L+w+perimiter of semi circle.

and the area is esentialy the area of rectangle +area of semi circle
and the area of the rectangle is L*W
and the area of the semi circle is _____________ (what we are trying to figure out)
so the formula for the area of a circle is pi*r^2
and r=(w/2)
so area of semi circle is pi(W/2)^2


so that means A=L*W+pi(W/2)^2


but they want it in terms of x (a single variable...however you have L and w in there which is 2 variables...so how would you make that one variable?)
*hint* has somethign to do with what's given...that the perimitr is 30...which is 2L+w+the perimiter of the arc of the semi circle...
and so what i was doing earlier was trying to get you to create an equation that relates the L and W to the perimiter of the semi circle
 
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  • #18
All I was given is the perimeter of the norman window, which is a recatangle surmounted by a semicircle, is equal to 30. Express the area of the window as a function of the width of the window. Sorry, the description I gave originally is straight from the calculus text. I got pir for the circumference of the semicircle.
 
  • #19
starchild75 said:
All I was given is the perimeter of the norman window, which is a recatangle surmounted by a semicircle, is equal to 30. Express the area of the window as a function of the width of the window. Sorry, the description I gave originally is straight from the calculus text. I got pir for the circumference of the semicircle.
Exactly. The window has one base, of length x, two heights of length y, and a semicircle of length [itex]\pi r[/itex]. Have you drawn a picture? Can you see what r is in terms of x? You know that the total length is 30. What equation is that?

Now, what is the area of an x by y rectangle? What is the area of the semi circle?
 
  • #20
H=(30-pi(w/2)-w)/2

A(w)=w(30-pi(w/2)-w)/2+.5pi(w/2)^2
 

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