# Proportion of light in a window

• December 10th 2009, 08:08 PM
Ife
Proportion of light in a window
A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only one-fourth as much light per unit area as clear glass does. The total perimeter is fixed. FInd the proportions of the window that will admit the most light. Neglect the thickness of the frame.

Now i am thinking the solution to this is something along the lines of:
$\frac {1}{2} \Pi D = \frac{1}{4} D x B$ (length by breath, where the length is equal to the diameter)
When this is simplified, i am getting $\frac{1}{4} \Pi : \frac{1}{2}B$

Is that a logical process to approach this??
• December 11th 2009, 02:20 AM
Hello Ife
Quote:

Originally Posted by Ife
A window is in the form of a rectangle surmounted by a semicircle. The rectangle is of clear glass, whereas the semicircle is of tinted glass that transmits only one-fourth as much light per unit area as clear glass does. The total perimeter is fixed. FInd the proportions of the window that will admit the most light. Neglect the thickness of the frame.

Now i am thinking the solution to this is something along the lines of:
$\frac {1}{2} \Pi D = \frac{1}{4} D x B$ (length by breath, where the length is equal to the diameter)
When this is simplified, i am getting $\frac{1}{4} \Pi : \frac{1}{2}B$

Is that a logical process to approach this??

Thanks for showing us your thinking. But you're approaching this in the wrong way. You need to set up some variables to represent the height and radius, and then represent the conditions in the question using these variables. I'll start you off.

Suppose that the height of the rectangle is $h$, and the radius of the semi-circle is $r$. Then, the width of the rectangle is $2r$. So the total perimeter is:
$2h+2r+\pi r = p$, say, where $p$ is a constant.

$\Rightarrow h = \tfrac12(p-[\pi+2]r)$ (1)
Now consider the areas. The area of the rectangle is:
$2rh = r(p-[\pi+2]r)$, from equation (1)
and the area of the semicircle is:
$\tfrac12\pi r^2$
Now suppose that the glass in the semi-circle admits $1$ unit of light per unit of area. Then the rectangle glass admits $4$ units of light per unit area. So the total light admitted, $L$, is given by:
$L = \tfrac12\pi r^2+4r(p-[\pi+2]r)$
OK. So you now need to:

• Simplify this expression

• Differentiate with respect to $r$, and equate the result to zero

• Solve the resulting equation, to get $r$ in terms of $p$

• Check that this value of $r$ gives a maximum value of $L$

• Substitute into (1) to get $h$ in terms of $p$

• Write down and simplify the ratio $h:2r$ to get the proportions of the rectangle when the most light is admitted

I make the ratio $h:2r = (8+3\pi):16$. Do you agree?

• December 11th 2009, 05:49 PM
Ife
Quote:

Hello IfeThanks for showing us your thinking. But you're approaching this in the wrong way. You need to set up some variables to represent the height and radius, and then represent the conditions in the question using these variables. I'll start you off.

Suppose that the height of the rectangle is $h$, and the radius of the semi-circle is $r$. Then, the width of the rectangle is $2r$. So the total perimeter is:
$2h+2r+\pi r = p$, say, where $p$ is a constant.

$\Rightarrow h = \tfrac12(p-[\pi+2]r)$ (1)
Now consider the areas. The area of the rectangle is:
$2rh = r(p-[\pi+2]r)$, from equation (1)
and the area of the semicircle is:
$\tfrac12\pi r^2$
Now suppose that the glass in the semi-circle admits $1$ unit of light per unit of area. Then the rectangle glass admits $4$ units of light per unit area. So the total light admitted, $L$, is given by:
$L = \tfrac12\pi r^2+4r(p-[\pi+2]r)$
OK. So you now need to:

• Simplify this expression

• Differentiate with respect to $r$, and equate the result to zero

• Solve the resulting equation, to get $r$ in terms of $p$

• Check that this value of $r$ gives a maximum value of $L$

• Substitute into (1) to get $h$ in terms of $p$

• Write down and simplify the ratio $h:2r$ to get the proportions of the rectangle when the most light is admitted

I make the ratio $h:2r = (8+3\pi):16$. Do you agree?

I agree. But can you tell me one thing please? How do i check the value to see that it gives a maximum? (After solving for r in terms of p). I've got to this part: $r= \frac{-(4p+9 \pi}{16}$
• December 11th 2009, 10:04 PM
Hello Ife
Quote:

Originally Posted by Ife
I agree. But can you tell me one thing please? How do i check the value to see that it gives a maximum? (After solving for r in terms of p). I've got to this part: $r= \frac{-(4p+9 \pi}{16}$

Differentiate a second time, and show that the second derivative is negative.

• December 13th 2009, 07:07 AM
Ife
Quote:

Hello IfeDifferentiate a second time, and show that the second derivative is negative.

Ok, i am continuing. Why i am finding the proportions of the rectangle to be h: 2r? (and i didn't understand that calculation you used to get that either). Shouldn't I be using the ratio of the rectangle:semicircle? Or do I need to give an answer of H by L by circumference of semicircle?

So far i have gotten to substitution of h in terms of p, and I am getting an answer of $h = \frac{1}{2}P - \frac{4P( \pi+2)}{14\pi+16}$
• December 13th 2009, 07:49 AM
Hello Ife
Quote:

Originally Posted by Ife
Ok, i am continuing. Why i am finding the proportions of the rectangle to be h: 2r? (and i didn't understand that calculation you used to get that either). Shouldn't I be using the ratio of the rectangle:semicircle? Or do I need to give an answer of H by L by circumference of semicircle?

So far i have gotten to substitution of h in terms of p, and I am getting an answer of $h = \frac{1}{2}P - \frac{4P( \pi+2)}{14\pi+16}$

Why $h:2r$? Well, the question you posed asked for the 'proportions' of the window. This is a bit vague. I just chose the height of the rectangle compared to its width (and therefore the diameter of the semi-circle). But you can choose a different description if you like.

Here's how I got the value of $r$ for the maximum value of $L$.

When we simplify
$L = \tfrac12\pi r^2+4r(p-[\pi+2]r)$ we get:
$L = 4pr-r^2(\tfrac72\pi+8)$

$\Rightarrow \frac{dL}{dr}= 4p - 2r(\tfrac72\pi+8)$
$=0$ when
$r=\frac{4p}{7\pi+16}$
When you substitute back for $h$ you get:
$h=\tfrac12\left(p-\frac{4p(\pi+2)}{7\pi+16}\right)$
which simplifies to:
$h=\frac{p(8+3\pi)}{2(7\pi+16)}$
and hence the answer I gave for the ratio $h:2r= (8+3\pi):16$.

• December 13th 2009, 08:05 AM
Ife
Quote:

Hello IfeWhy $h:2r$? Well, the question you posed asked for the 'proportions' of the window. This is a bit vague. I just chose the height of the rectangle compared to its width (and therefore the diameter of the semi-circle). But you can choose a different description if you like.

Here's how I got the value of $r$ for the maximum value of $L$.

When we simplify
$L = \tfrac12\pi r^2+4r(p-[\pi+2]r)$ we get:
$L = 4pr-r^2(\tfrac72\pi+8)$

$\Rightarrow \frac{dL}{dr}= 4p - 2r(\tfrac72\pi+8)$
$=0$ when
$r=\frac{4p}{7\pi+16}$
When you substitute back for $h$ you get:
$h=\tfrac12\left(p-\frac{4p(\pi+2)}{7\pi+16}\right)$
which simplifies to:
$h=\frac{p(8+3\pi)}{2(7\pi+16)}$
and hence the answer I gave for the ratio $h:2r= (8+3\pi):16$.