a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular window. Find the dimensions of a Norman window of maximum area it the total perimiter is 52 feet.
To solve this problem, we need to find the dimensions of the Norman window that maximize its area, while considering the given total perimeter of 52 feet.
Let's start by breaking down the problem into smaller parts:
1. Assign variables:
- Let x be the width of the rectangular window.
- Let y be the height of the rectangular window.
- Let r be the radius of the semicircle.
2. Calculate the perimeters:
- The perimeter of the rectangular part is: 2x + 2y.
- The perimeter of the semicircle (half of the circumference) is: 1/2 * π * r.
Since the total perimeter is given as 52 feet, we can write the equation:
2x + 2y + 1/2 * π * r = 52
3. Calculate the area:
- The area of the rectangular part is: x * y.
- The area of the semicircle is: 1/2 * π * r^2.
The total area, A, is the sum of the area of the rectangular part and the semicircle:
A = x * y + 1/2 * π * r^2
To maximize the area, we need to find the maximum values of x, y, and r under the constraint of the given total perimeter.
Now, let's solve the problem step by step:
1. Rearrange the equation for the perimeter using the given values:
2x + 2y + 1/2 * π * r = 52
2. Solve for x:
2x = 52 - 2y - 1/2 * π * r
x = (52 - 2y - 1/2 * π * r) / 2
3. Substitute the expression for x into the area equation:
A = (52 - 2y - 1/2 * π * r) / 2 * y + 1/2 * π * r^2
4. Simplify the equation:
A = 26y - y^2 + 1/4 * π * r^2 - 1/2 * π * r * y
5. To find the maximum value of A, we take the derivative of A with respect to y and set it equal to zero:
dA/dy = 26 - 2y - 1/2 * π * r = 0
6. Rearrange the equation to solve for r:
r = (52 - 4y) / π
7. Substitute the value for r back into the equation for y:
26 - 2y - 1/2 * π * (52 - 4y) / π = 0
8. Simplify the equation:
26 - 2y - 26 + 2y - 2y = 0
0 = 0
9. Since we get 0 = 0, it means that for any value of y, the equation holds true. This implies that there is no maximum area, and the shape can have infinitely large dimensions, which is not physically possible.
Therefore, there is no solution for a Norman window of maximum area under the given constraints.