Problem 26 - Calculus Early Transcendentals

Question

A church window consisting of a rectangle topped by a semi circle is to have a perimeter \(p,\) Find the radius of the semicircle if the area of the window is to be maximum.

Step-by-Step Solution

Answer

The semicircle radius for maximum area is \( \frac{p}{4 + \pi} \).

Step 1: Define Variables and Formulas

Let's define the necessary variables. Let \( r \) be the radius of the semicircle, and \( w \) be the width of the rectangle (which is equal to the diameter of the semicircle, \( 2r \)). Let \( h \) be the height of the rectangle. The perimeter of the window is given by \( p = 2h + 2r + \pi r \), and the area \( A \) is given by the area of the rectangle plus the area of the semicircle: \( A = 2rh + \frac{1}{2}\pi r^2 \).

Step 2: Express Height in Terms of Other Variables

From the perimeter equation, solve for the height \( h \): \[ h = \frac{p - 2r - \pi r}{2}. \] Substitute this expression for \( h \) into the area formula so that the area \( A \) becomes a function of \( r \) alone.

Step 3: Substitute Height into Area Equation

Substituting \( h \) into the area formula, we have: \[ A = 2r\left(\frac{p - 2r - \pi r}{2}\right) + \frac{1}{2}\pi r^2. \] Simplify this equation: \[ A = r(p - 2r - \pi r) + \frac{1}{2}\pi r^2. \] Expand and combine terms to get the quadratic function of \( r \): \[ A = pr - 2r^2 - \pi r^2 + \frac{1}{2}\pi r^2. \] Simplify to \[ A = pr - 2r^2 - \frac{1}{2}\pi r^2. \]

Step 4: Find the Derivative of the Area Function

To find the value of \( r \) that maximizes the area, take the derivative of the area function with respect to \( r \): \[ \frac{dA}{dr} = p - 4r - \pi r. \]

Step 5: Set the Derivative to Zero and Solve

Set the derivative equal to zero to find the critical points: \[ p - 4r - \pi r = 0. \] This simplifies to \[ p = r(4 + \pi). \] Solve for \( r \): \[ r = \frac{p}{4 + \pi}. \]

Step 6: Verify Maximum Conditions

To confirm this critical point is indeed a maximum, check the second derivative \( \frac{d^2A}{dr^2} \): \[ \frac{d^2A}{dr^2} = -4 - \pi. \] Since this is negative, the critical point provides a maximum. Therefore, the radius that maximizes the area is \( \frac{p}{4 + \pi} \).

Key Concepts

Perimeter and Area

When dealing with shapes, particularly those that incorporate both straight lines and curves, understanding perimeter and area is crucial in optimization problems in calculus.
The perimeter refers to the total distance around the shape. In our exercise, the window shape is a combination of a rectangle and a semi-circle. The perimeter for this window is the sum of the perimeters of the rectangle and the semicircle:

• The rectangle's width is the same as the diameter of the semicircle, so its perimeter is twice its height plus the width.
• The semicircle's perimeter includes its arc (half of the circumference of a full circle with radius \( r \), which is \( \pi r \)).

Together, they form the equation \( p = 2h + 2r + \pi r \), where \( p \) represents the total perimeter.
The area is the total region enclosed within the shape. For the window, this area is the sum of the area of the rectangle and the semicircle:

• The rectangle's area is \( 2rh \), where \( r \) is the radius of the semicircle and hence the width of the rectangle.
• The semicircle's area is half the area of a full circle, \( \frac{1}{2}\pi r^2 \).

These measurements must be rearranged to use the given perimeter to express the shape's dimensions in terms solely of \( r \) to help find the maximum area.

Critical Points

Critical points are those values of the variable where the function's rate of change, indicated by its derivative, equals zero or is undefined. In the context of our exercise, we seek to find the radius of the semicircle, \( r \), at which the area of the window is maximized. These are key to finding either a maximum or minimum value of a function.
The derivative of the area function \( A \) with respect to \( r \) is calculated first to find these critical points. We rearranged the area formula in terms of \( r \), and then took its derivative:

• The critical point equation is given by setting \( \frac{dA}{dr} = p - 4r - \pi r \) to zero.

This gives us the opportunity to simplify and solve for \( r \), which indicates at which point the area does not increase or decrease but is at its peak (a maximum in our case). These turning points are where the optimization occurs.

Derivative and Second Derivative Test

The derivative test, including the second derivative test, is a tool used to determine the nature of critical points found for a function—indicating whether they are maxima, minima, or points of inflection.
In the exercise, once the first derivative \( \frac{dA}{dr} \) is set to zero to find the critical point, checking whether this critical point is a maximum involves the second derivative.

• The second derivative \( \frac{d^2A}{dr^2} \) is calculated as \( -4 - \pi \).
• If this second derivative is negative, it tells us that the concavity of the function is downwards, verifying that the critical point provides a maximum area.

Hence, in our window exercise, since the second derivative \( \frac{d^2A}{dr^2} \) turned out to be negative, \( r = \frac{p}{4 + \pi} \) is a local maximum. This showcases the usefulness of second derivatives in confirming that the simplified value derived indeed maximizes the area for the given perimeter.