Building windows: Window World, Inc., is responsible for designing new windows for the expansion of the campus chapel. The current design is shown in the figure. The metal trim used to secure the perimeter of the frame is 126″ long. If the maximum window area is desired (to let in the most sunlight), what will be (a) the dimensions of the rectangular portion of each window? (b) the total area of each window
no figure, but from the context it appears to be a rectangle and something else. Since it's a church, and the same kind of problem has appeared here before, I assume the top portion is a semi-circle.
So, with width w and height h, the perimeter is w+2h+pi*w/2. The trim is not used between the top and bottom portions of the window.
So, we have
w+2h+pi*w/2 = 126
so,
h = 63 - (pi+2)w/4
window area is rectangle + half-circle
a = wh + pi/8 w^2
= w(63 - (pi+2)w/4) + pi/8 w^2
= 63w - pi/8 (2pi+3)w^2
da/dw = 63 - pi/4 (2pi+3)w
max/min area is where da/dw = 0
w = 252/(3pi + 2pi^2) = 8.64
Now you can figure the areas.
To find the dimensions of the rectangular portion of each window, we need to make some assumptions and solve for the variables.
Let's assume the rectangular portion of each window has a length of L and a width of W.
We know that the metal trim used to secure the perimeter of the frame is 126 inches long. Since there are four sides to the rectangular portion of the window, the total length of the sides (perimeter) can be expressed as:
Perimeter = 2L + 2W
Given that the perimeter is 126 inches, we can set up the following equation:
2L + 2W = 126
Now, let's solve for L in terms of W:
2L = 126 - 2W
L = (126 - 2W) / 2
L = 63 - W
Now, we have the length L in terms of the width W. To maximize the window area, we want to maximize the rectangular portion's dimensions. Since both the length and width must be positive, we can say that W > 0 and L > 0.
To find the maximum window area, we need to multiply the length L by the width W:
Window Area = L * W
Window Area = (63 - W) * W
Window Area = 63W - W^2
Now, to find the value of W that maximizes the area, we can take the derivative of the Window Area equation with respect to W and solve for W:
d(Window Area)/dW = 63 - 2W
Setting the derivative equal to 0 and solving for W:
63 - 2W = 0
2W = 63
W = 31.5
Since W represents the width, it cannot be negative. Therefore, we can ignore the negative value and consider W = 31.5 inches as the width of the rectangular portion of each window.
Now, substitute the value of W back into the equation for L:
L = 63 - W
L = 63 - 31.5
L = 31.5
Therefore, the dimensions of the rectangular portion of each window are 31.5 inches (width) by 31.5 inches (length).
To find the total area of each window, simply multiply the width and length:
Total Area = L * W
Total Area = 31.5 * 31.5
Total Area = 992.25 square inches
So, the total area of each window will be 992.25 square inches.