Math Problem Statement

  1. A Norman window has the shape of a rectangle surmounted by a semicircle. (The diameter of the semicircle is equal to the width of the rectangle. See the figure below.) If the perimeter of the window is 32 ft, find the value of x so that the greatest possible amount of light is admitted.

Solution

To maximize the amount of light admitted through a Norman window, we need to maximize its area, given the perimeter constraint. Let's break down the problem and solve for xx, the width of the rectangle (and diameter of the semicircle) that will maximize the area.

Step 1: Define Variables

  1. Let xx be the width of the rectangular part of the window (and thus the diameter of the semicircle).
  2. Let hh be the height of the rectangular part of the window.
  3. The perimeter constraint is 32 ft.

Step 2: Express Perimeter in Terms of xx and hh

The perimeter of the Norman window includes:

  • The two vertical sides of the rectangle, each of height hh,
  • The width xx of the rectangle, and
  • The semicircular arc on top, which has a radius of x2\frac{x}{2} and thus a circumference of πx2\pi \frac{x}{2}.

So, the perimeter PP is: P=2h+x+πx2P = 2h + x + \frac{\pi x}{2} Since the perimeter is given as 32 ft, we set up the equation: 2h+x+πx2=322h + x + \frac{\pi x}{2} = 32

Step 3: Solve for hh in Terms of xx

Rearrange the equation to isolate hh: 2h=32xπx22h = 32 - x - \frac{\pi x}{2} h=32xπx22h = \frac{32 - x - \frac{\pi x}{2}}{2} h=16x2πx4h = 16 - \frac{x}{2} - \frac{\pi x}{4}

Step 4: Express the Area in Terms of xx Only

The total area AA of the window is the sum of the area of the rectangle and the area of the semicircle.

  1. The area of the rectangle is: Area of rectangle=xh\text{Area of rectangle} = x \cdot h
  2. The area of the semicircle is: Area of semicircle=12π(x2)2=πx28\text{Area of semicircle} = \frac{1}{2} \pi \left(\frac{x}{2}\right)^2 = \frac{\pi x^2}{8}

Thus, the total area AA is: A=xh+πx28A = x \cdot h + \frac{\pi x^2}{8}

Substitute h=16x2πx4h = 16 - \frac{x}{2} - \frac{\pi x}{4}: A=x(16x2πx4)+πx28A = x \left(16 - \frac{x}{2} - \frac{\pi x}{4}\right) + \frac{\pi x^2}{8} Simplify this expression to find AA as a function of xx alone: A=16xx22πx24+πx28A = 16x - \frac{x^2}{2} - \frac{\pi x^2}{4} + \frac{\pi x^2}{8} Combine terms involving x2x^2: A=16xx2(12+π4π8)A = 16x - x^2 \left(\frac{1}{2} + \frac{\pi}{4} - \frac{\pi}{8}\right) Calculate the constants and simplify further to maximize A(x)A(x) by finding the derivative and setting it to zero.

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Math Problem Analysis

Mathematical Concepts

Geometry
Optimization
Calculus

Formulas

Perimeter P = 2h + x + (πx/2)
Area A = x * h + (πx^2/8)

Theorems

Optimization in calculus

Suitable Grade Level

Grades 11-12