Question

a norman window is a window with a semicircle on top of a regular rectangular window as shown in the diagram.what should the dimensions of the rectangular part of the norman window be to allow in as much light as possible if there is only 12 ft of framing material available

Answer 1

To maximize the area of a Norman window with 12 feet of framing material, we must use calculus to derive the optimal dimensions of the rectangle while considering the perimeter constraints.

Step-by-step explanation:

The student is asking about the optimal dimensions for a Norman window with 12 feet of framing material, which is a combination of a rectangle and a semicircle on its top. To maximize the area of the window, which translates to allowing in as much light as possible, we need to use calculus to find the dimensions of the rectangle given the constraint on the perimeter (or framing material).

Let's denote the width of the rectangular part as w and the height as h. Since it's a semicircle on top of the rectangle, the diameter of the semicircle is equal to the width of the rectangle. The total perimeter of the window using the 12 feet of framing material must include the two sides and bottom of the rectangle plus half the circumference of the semicircle (since it's a semicircle, not a full circle). The formula for the perimeter (P) is:

P = 2h + w + (π * w)/2

Since we have 12 feet of material: 2h + w + (π * w)/2 = 12. The area (A) of the window is the sum of the area of the rectangle and the semicircle: A = w * h + (π * w2)/8. To maximize the area, we take the derivative of A with respect to w, set it to zero, and solve for w. Then, we solve for h using the perimeter equation given w.

Optimization Steps:

1. Write the perimeter constraint equation: 2h + w + (π * w)/2 = 12.
2. Express h in terms of w from the constraint equation.
3. Write the area equation: A = w * h + (π * w2)/8
4. Substitute h from step 2 into the area equation to get A as a function of w only.
5. Take the derivative of A with respect to w, set it to zero and solve for w to find the width that maximizes the area.
6. Substitute the value of w back into the constraint equation to find h.
7. Verify that the second derivative test for A is negative, confirming a maximum area.

The dimensions that maximize the area under the given constraints will allow in the most light possible.