Designing a Window

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
For your house you are designing a new window which will comprise a semi circle surmounted onto a rectangle. The total perimeter is to be 10 metres exactly. What are the dimensions of the window if the design will allow for the most light to enter the room via the greatest possible surface area? What is the total area that is achieved?

Ron.

( I know that a lot of people are enjoying working through these problems that I post, especially Richard, so I hope you will all enjoy this one too! )
 
Any thoughts as yet as to how this one might be solved?

Richard,...how are you going with this one so far?

Ron.
 
darth sidious said:
SydneyRoverP6B said:
I'll post the solution over the weekend.

Ron.
Click to expand...

No!!! I want to have a try myself first! :)
Click to expand...

Go for it Darth!

I wont't let on what I think

2.8

Ooops :oops:

I reckon, purely by intuition, that it will be a semi circle on a semi square, i.e. A figure of 2.8M high and 2.8M wide gives a perimeter of 10M and and area of 7M2
 
Hello Richard,

You have the correct area,..it is indeed 7 square meters!.. :D Now without using intuition, how do you prove that to be the correct area?

Ron.
 
The area is (Pi * r2)/2 added to 2*r2

r = 1.4 if the diameter is 2.8

(3.142*1.96)/2 = 3.08 (semi circle area)

2*1.4*1.4 = 3.92 (half square area)

3.08 + 3.92 = 7

Perimter = Pi*d/2 added to d*4/2

= 4.4 + 5.6 = 10
 
Hello Richard,

Your mathematics is correct using a radius of r = 1.4. How though do you prove that those measurements will deliver the greatest possible surface area?

I'll post another question which will extend what you have already achieved.

Very well done... :D

Ron.
 
SydneyRoverP6B said:
Hello Darth,

How are you going with it? Richard's intuition was right, but it needs to be proven.

Ron.
Click to expand...

Not that far! :( (Had a bad case of stomach ache last night, which didn't help a bit!)

Show us the solution, I'm very interested in seeing how this is done! :D
 
Hello Darth,

I hope you are feeling better now!

I am pleased that you are really interested in how it is done,.. :D I'll write up the solution now.

Ron.
 
Here is the formal solution to this problem.

As the only information provided is the total perimeter of 10 metres, so an expression must be written involving initially ‘r’ and ‘h’ as the two unknowns. I have chosen ‘r’ for radius and ‘h’ for height of the rectangle.

As the perimeter is a constant only the area can vary, so the expression for perimeter is…

10 = 2r + 2h + PIr

Now as this equation has two unknowns, rewrite it in terms of ‘r’

2h = 10 – 2r – PIr
= 10 – r(2 + PI)
h = ((10 – r(2 + h))/2 [ 'h' is now a function of 'r'] so i'll write...
h(r) = ((10 - r(2 + h))/2

The area of the semi circle is Pir^2/2 and that of the rectangle 2rh

Substituting in the above expression for ‘h’ the total area expressed as a function of ‘r’ is…

A(r) = Pir + 10r – r^2(2 + PI)

Now in order to prove that the area obtained will be the maximum possible for the given perimeter of 10 metres, we must first take the derivative dA/dr of A(r) to obtain A’(r)

A’(r) = Pir + 10 – 2r(2 + PI)

Now to obtain the critical point set A’(r) equal to zero and solve for ‘r’

r = 10/(4 + PI)

To prove that this length is indeed the maximum possible length that will give the greatest possible surface area, and remember that ‘h’ is also expressed as a function of ‘r’ test points either side of ‘r’ using the first derivative test. For values less than ‘r’ deliver a positive outcome whilst those larger than ‘r’ a negative outcome. Thus the value obtained is proven to be the maximum within the closed interval [10/(2 + PI) , 5]. These values provide the domain of ‘r’ and upon inspection the end points offer degenerate areas.

From the value of ‘r’ the length of ‘h’ is obtained by substituting back into the above equation for ‘h’ which also gives 10/(4 + PI)

So the rectangle has a width twice its height and the area obtained by the calculated values is the maximum possible within the constraint of a constant perimeter.

Any item on any scale can be designed to meet any requirments using Calculus....the mathematics of continuous change.

The problems I have posted so far do allow for an alternative methodology as used so successfully by Richard, but as the problems become more difficult, so the aternatives become fewer.

Ron.
 
SydneyRoverP6B said:
Hello Darth,

I hope you are feeling better now!

I am pleased that you are really interested in how it is done,.. :D I'll write up the solution now.

Ron.
Click to expand...

I am much better thank you!

I have actually practised my Excel work and done it numerically. It does seem that 7 is the largest possible area. Numerically differentiating about r=1.4 does yield a <very close to zero> value, and mentally differentiating the numerical derivative yield a -ve result, which suggests a maximum at r=1.4

I'm starting to enjoy numerical methods, to be honest!

Addendum: The numerical list was not keeping the format when posted here! :( Oh well!

I see how you got the answer, and even did it on paper as reading it. That's very good! Oh, and I got the same answer too! I actually differentiated twice, and found the second derivative was < 0 {-(4+pi), to be precise!}, so the maximum occurs at r = 10/(4 + pi) metres!
 
Hello Darth,

How do you like the 'classical' approach?

I have never tired numerical methods on excel.

I do all my calculations using my calculator or by hand.

Ron.
 
SydneyRoverP6B said:
Hello Darth,

How do you like the 'classical' approach?

I have never tired numerical methods on excel.

I do all my calculations using my calculator or by hand.

Ron.
Click to expand...

Excellent, it was very interesting. I do like to check with using numerical also, just to prove to myself I got it right! I think Quattro also likes to check numerically too, seeing he used that method with the balloon and P6 problem.

To be fair, I do try to use mental arithmetic also. It never ceases to amaze me how many people nowadays seem to be unable to do quick simple-ish calcs in their head! Mind you, even I am guilty of that sometimes!
 
I am glad you liked it,..and yes the second derivative being less than zero...concave down so a maximum value.

I try and do as much as I can without using my calculator too....keeps the mind alive.

How do you like the cone problem?

Ron.
 
SydneyRoverP6B said:
I am glad you liked it,..and yes the second derivative being less than zero...concave down so a maximum value.

I try and do as much as I can without using my calculator too....keeps the mind alive.

How do you like the cone problem?

Ron.
Click to expand...

Didn't see the cone problem! :-S

Have taken a stab at it. I think I've gotten it right... at least I hope I have!
 
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