Math Problem Statement

A Norman window consists of a semicircle adjoining a rectangular window. If the perimeter of the Norman window is 23 ft, find the dimensions of r (the radius of the semicircle) and w (the height of the rectangle) that allow the window to admit the most light.

Solution

To solve the given problem involving the Norman window, let's break it down step-by-step.

1. Understanding the components:

  • The window consists of:
    1. A semicircle at the top with radius rr.
    2. A rectangle below the semicircle, with:
      • Width 2r2r (since the width of the rectangle equals the diameter of the semicircle).
      • Height ww (denoted in the figure).

2. Perimeter constraint:

The perimeter of the Norman window includes:

  • The semicircle's curved edge: πr\pi r (half of the circumference of a full circle).
  • The two vertical sides of the rectangle: 2w2w.
  • The bottom side of the rectangle: 2r2r (same as the diameter of the semicircle).

Thus, the total perimeter is: πr+2w+2r=23ft.\pi r + 2w + 2r = 23 \, \text{ft}.

3. Optimization goal:

We need to maximize the area to allow the most light. The total area consists of:

  • Area of the semicircle: 12πr2\frac{1}{2} \pi r^2.
  • Area of the rectangle: 2rw2r \cdot w.

Thus, the total area is: A(r,w)=12πr2+2rw.A(r, w) = \frac{1}{2} \pi r^2 + 2r w.

4. Solve for ww in terms of rr using the perimeter:

From the perimeter equation: πr+2w+2r=23.\pi r + 2w + 2r = 23. Rearrange to solve for ww: 2w=23πr2r,2w = 23 - \pi r - 2r, w=23πr2r2.w = \frac{23 - \pi r - 2r}{2}.

5. Express the area in terms of rr only:

Substitute the expression for ww into the area formula: A(r)=12πr2+2r(23πr2r2).A(r) = \frac{1}{2} \pi r^2 + 2r \left( \frac{23 - \pi r - 2r}{2} \right).

Simplify: A(r)=12πr2+r(23πr2r).A(r) = \frac{1}{2} \pi r^2 + r(23 - \pi r - 2r). A(r)=12πr2+23rπr22r2.A(r) = \frac{1}{2} \pi r^2 + 23r - \pi r^2 - 2r^2. A(r)=12πr22r2+23r.A(r) = -\frac{1}{2} \pi r^2 - 2r^2 + 23r.

6. Maximize the area:

To find the value of rr that maximizes the area, take the derivative of A(r)A(r) with respect to rr and set it equal to zero: dAdr=πr4r+23=0.\frac{dA}{dr} = -\pi r - 4r + 23 = 0. (π+4)r+23=0.-(\pi + 4)r + 23 = 0. (π+4)r=23.(\pi + 4)r = 23. r=23π+4.r = \frac{23}{\pi + 4}.

7. Calculate the corresponding ww:

Now that we have rr, substitute it into the equation for ww: w=23πr2r2.w = \frac{23 - \pi r - 2r}{2}.

Using r=23π+4r = \frac{23}{\pi + 4}, you can substitute this value into the expression to compute ww.


Would you like me to proceed with the numerical calculation for these values or clarify any step?

5 Related Questions:

  1. How do we ensure this critical point yields the maximum area?
  2. What is the area of the Norman window with these dimensions?
  3. How does the perimeter constraint limit the design?
  4. What happens if the perimeter was increased or decreased?
  5. Can calculus guarantee the most optimal dimensions in all cases?

Tip:

Always verify your derivative calculations to ensure you correctly identify critical points for optimization problems.

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Math Problem Analysis

Mathematical Concepts

Calculus
Optimization
Geometry

Formulas

Perimeter: πr + 2w + 2r = 23 ft
Area of semicircle: (1/2)πr^2
Area of rectangle: 2r * w
Total area: (1/2)πr^2 + 2r * w

Theorems

First Derivative Test for Optimization
Geometry of Circles and Rectangles

Suitable Grade Level

Grade 11-12, Early College