Light through a window

**Anthony Hodson** · December 9th 2005, 03:22 AM

I have a problem that I have solution to but not the proof. Can anyone help?

The problem is this:

A window, shaped as a rectangle with a semicircle on top, has a perimeter of 5m. The semicircle has a radius r and the rectangle has sides which are effectively 2r x h (height) what value of r gives the maximum light (area).

Simple question I know, but have problems with the details.

**CaptainBlack** · December 9th 2005, 04:39 AM

Originally Posted by Anthony Hodson

I have a problem that I have solution to but not the proof. Can anyone help?

The problem is this:

A window, shaped as a rectangle with a semicircle on top, has a perimeter of 5m. The semicircle has a radius r and the rectangle has sides which are effectively 2r x h (height) what value of r gives the maximum light (area).

Simple question I know, but have problems with the details.

The perimeter is:

$2h\ +\ 2r\ +\ \pi r\ =\ 5$

rearrange this to give $h$ in terms of $r$ .

The area is:

$Area\ =\ 2r.h\ +\ \pi .r^2/2$ .

Substitute the representation you have for $h$ in terms of $r$ into
this, then find the $r$ that maximises the area.

You do this by differentiating $Area$ w.r.t. $r$ and
setting the derivative to zero, and solving for $r$ .

RonL

**Anthony Hodson** · December 9th 2005, 04:50 AM

Thanks,

I have got that far, although I see you have 2*pi*r for the perimeter, although i used pi*r (semicircle). However, Its the last part I can't get to. I cant see the max and min values of r?

can you help?

**CaptainBlack** · December 9th 2005, 05:45 AM

Originally Posted by Anthony Hodson

Thanks,

I have got that far, although I see you have 2*pi*r for the perimeter, although i used pi*r (semicircle).

Opps, typo, you are right

However, Its the last part I can't get to. I cant see the max and min values of r?

can you help?

**CaptainBlack** · December 9th 2005, 05:53 AM

Originally Posted by Anthony Hodson

Thanks,

Its the last part I can't get to. I cant see the max and min values of r?

can you help?

We have <you will need to check this - NOT GUARANTEED TYPO FREE >:<Corrected Now - I hope>

$h\ =\ \frac{5-r(2+\pi)}{2}$ ,

so substituting the RHS of this for $h$ in the $Area$
equation gives:

$Area\ =\ 5r\ -\ (2+\pi/2)r^2$

Then:

$\frac{d\ Area}{dr}\ =\ 5\ -\ 2(2+\pi/2)r$ .

Now set

$\frac{d\ Area}{dr}\ =\ 0$

and solve for $r$

That is solve:

$5\ -\ 2(2+\pi/2)r\ =\ 0$

for $r$ .

RonL

**Anthony Hodson** · December 9th 2005, 06:25 AM

Thanks again, brilliant response,

not sure if its a typo or not, your area equation seems to be missing a semicircle?

It would appear i am on the right track, however I am expecting two values for r = 0.707 and a - figure?

I am sure that this a simple equation, but I must be making silly mistakes, I am still thinking that the Derivative should a quadratic equation but I may be wrong?

Regards
Tony

**CaptainBlack** · December 9th 2005, 07:28 AM

Originally Posted by Anthony Hodson

Thanks again, brilliant response,

not sure if its a typo or not, your area equation seems to be missing a semicircle?

$<br /> Area\ =\ 2r.h\ +\ \pi .r^2/2<br />$

First term is the area of the rectangular part, the second the
area of the semi-circle.

$h\ =\ \frac{5-r(2+\pi)}{2}$

so,

$Area\ =\ 2r.\frac{5-r(2+\pi)}{2}\ +\ \pi .r^2/2$ ,

$Area\ =\ r.5-r^2(2+\pi)\ +\ \pi .r^2/2$ ,

$Area\ =\ r.5-2.r^2-\pi .r^2/2$ ,

$Area\ =\ r.5-r^2.(2+\pi/2)$

which might now be right

I will correct the previous post when I get a chance.

RonL

**Anthony Hodson** · December 9th 2005, 08:01 AM

Many Thanks,

**ticbol** · December 9th 2005, 10:56 AM

Originally Posted by Anthony Hodson

I have a problem that I have solution to but not the proof. Can anyone help?

The problem is this:

A window, shaped as a rectangle with a semicircle on top, has a perimeter of 5m. The semicircle has a radius r and the rectangle has sides which are effectively 2r x h (height) what value of r gives the maximum light (area).

Simple question I know, but have problems with the details.

Here is one way.

Perimeter, P = (pi)r +h +2r +h = 5
2h +r(2+pi) = 5
Differentiate both sides with respect to r,
2(dh/dr) +(2+pi) = 0
dh/dr = -(2+pi)/2 -------------***

Area, A = pi(r^2)/2 +2r(h)
Differentiate both sides with respect to r,
dA/dr = (pi)r +2[r(dh/dr) +h]
Set dA/dr to zero, for max A, and plug in the value of dh/dr,
0 = (pi)r +2[r{-(2+pi)/2} +h]
0 = (pi)r -r(2+pi) +2h
0 = (pi)r -2r -(pi)r +2h
r = h ---------for max A.

Back to the perimeter,
2h +r(2+pi) = 5
Substitute r for h,
2r +r(2+pi) = 5
r(2 +2+pi) = 5
r = 5/(4+pi) = 0.70 m. -------for max A, answer.

**CaptainBlack** · December 9th 2005, 10:54 PM

Originally Posted by ticbol

Here is one way...

And here is another, which unfortunately will be all Greek to Antony
but I am posting anyway for a bit of fun

.

Maximise:

$Area(r,h)\ =\ 2r.h\ +\ \pi.r^2/2$ ,

subject to the constraint:

$Perimeter(r,h)\ =\ 2h\ +\ r(2+\pi)\ =\ 5$ .

We will use the Lagrange Multipliers, so we let:

$H(r,h,\lambda)\ =\ 2r.h\ +\ \pi.r^2/2\ +\ \lambda[2h\ +\ r(2+\pi)\ -\ 5]$

and we seek solutions of:

$\frac{\partial H}{\partial r}\ =\ 2h\ +\ \pi r\ +\ \lambda (2+\pi)\ =\ 0$

$\frac{\partial H}{\partial h}\ =\ 2r\ +\ 2 \lambda\ =\ 0$

(the constraint also must hold - it expressed the condition that
$\partial H/ \partial \lambda\ =\ 0$ ).

The second of these equations tells us that:

$\lambda \ =\ -r$ ,

which when substituted into the first of the equations gives:

$h\ =\ r$ ,

which when substituted back into the constraint gives:

$r\ =\ \frac{5}{4+\pi}$

RonL

**Anthony Hodson** · December 14th 2005, 06:55 AM

your right greek, well done to all.

Thanks

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