window is in the form

Question

window is in the form of a sequare surmounted by a semicircle.

The sequare of clear glass, whereas the semicircle is of tinted glass that

transmits only quarter 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

Answer 1

I am assuming that the window has a rectangular shape, not square. If the window was square, and there was a given perimeter, then we would not have to solve for the dimensions of the window because it could be solved just by knowing the perimeter.

Assume that x is the width of the rectangular portion window, and y is the height of the rectangular portion of the window. Because the semicircle, with diameter d, is surmounted onto the window, the width of the rectangular portion of the window is equal to the diameter of the semicircle portion, i.e. x = d.

1. x = window width = circle diameter
2. y = window height
3. r = x/2

Now we will solve for the perimeter of the window. The perimeter of the rectangular portion will be x + 2y (we exclude the top portion of the rectangular window, because it is continuous with the semicircular portion of the window), and the circular portion will be equal to πx/2 (which is half the circumference of a circle).

hence:

1. P = x + 2y + πx/2 = 2r + 2y + πr

Now we will solve for the light travelling through the window. We will assume the light trasnmitted per unit area of clear glass is 1, so the light transmitted per unit area of tinted glass is 1/4. The rectangular light will be equal to the area of the rectangular portion multiplied by the light constant of clear glass, i.e. 1*2ry = 2ry. The semicircular portion will be equal half the area of a circle multiplied by the light constant of tinted glass, (1/4)* πr²/2. Hence,

1. L = 2ry + πr²/8

Now, we will isolate y from the perimeter equation, and then plug that value of y into the light equation

1. y = (P - r(2 + π))/2
2. L = r(P - r(2 + π)) + πr²/8

Because P is constant, the only variables in the light equation are L and r. We can differentiate L with respect to r (i.e. dL/dr) and solve for when dL/dr = 0 to find the point at which the light transmitted is at a maximum.

1. dL/dr = P - 2r(2 + π) + πr/4
2. 0 = P - 2r(2 + π) + πr/4
3. r = 4P/(16 + 7π)

Finally, we can take this and plug into the above equations to find the final solutions

1. x = 2r = 8P/(16+ 7π)
2. y = P - r(2 + π)/2 [I neglected from writing taking the equation of P and inserting here because the equation could get unruly and long

Answers:

1. r = 4P/(16 + 7π)
2. x = 8P/(16+ 7π)
3. y = P - r(2 + π)/2