Problem 15 - Beginning and Intermediate Algebra

Question

Solve. The Soo family wants to fence in a rectangular area to hold their dogs. One side of the pen will be their barn. Find the dimensions of the pen of greatest area that can be enclosed with 48 ft of fencing.

Step-by-Step Solution

Answer

The dimensions of the pen of greatest area that can be enclosed with 48 ft of fencing are \(x = 24\) ft (length parallel to the barn) and \(y = 12\) ft (width perpendicular to the barn).

Step 1: Rewrite the constraint in terms of one variable

We have the constraint equation x + 2y = 48. We will rewrite it in terms of one variable (let's choose y) by solving for x: x = 48 - 2y - (1a)

Step 2: Substitute the constraint into the area equation

Now that we have x in terms of y (from equation (1a)), let's substitute that into the area equation (2) to get A as a function of y alone: A(y) = (48 - 2y)y A(y) = 48y - 2y^2

Step 3: Find the derivative of A(y) with respect to y

In order to find the maximum area, we need to find the critical points of A(y). To do this, we'll take the derivative of A(y) with respect to y: A'(y) = 48 - 4y

Step 4: Set A'(y) to zero and solve for critical points

To find critical points, we'll set A'(y) to 0 and solve for y: 0 = 48 - 4y y = 12

Step 5: Verify that we've found a maximum

To check that y = 12 ft corresponds to a maximum, we will examine A"(y), the second derivative of A(y) with respect to y. If A"(y) is less than 0, then we have a maximum. A"(y) = -4 Since A"(y) = -4 < 0, we have confirmed that y = 12 ft corresponds to a maximum.

Step 6: Find the corresponding value for x

Now that we have y = 12 ft, we can use the constraint equation (1a) to solve for x: x = 48 - 2(12) x = 24 ft The dimensions of the pen of greatest area that can be enclosed with 48 ft of fencing are x = 24 ft (length parallel to the barn) and y = 12 ft (width perpendicular to the barn).

Key Concepts

Rectilinear Shapes

Rectilinear shapes are shapes with straight sides and angles, commonly known as polygons. These include familiar shapes like rectangles, squares, and triangles. Rectangles, specifically, have four sides and four right angles. In many real-world applications, including fencing problems, rectangles are often used because of their simplicity and the efficiency in utilizing space.

When dealing with rectilinear shapes like rectangles, the perimeter is the sum of all the side lengths. For a rectangle to maximize area given a certain perimeter, like in the Soo family's situation with the 48 ft of fencing, you typically want the shape to be as square-like as possible given certain conditions like attaching one side to a barn.

Derivatives in Calculus

Derivatives in calculus help us understand how a function is changing at any given point. In optimization problems, like finding the maximum area of a pen with a specific amount of fencing, derivatives are our main tool. By differentiating the area function, we can find its slope at different points.

With the Soo family's fence problem, we differentiated the area with respect to one of the dimensions (y), producing a new function, A'(y). This derivative tells us the rate of change of the area as y changes. In simple terms, if A'(y) is positive, increasing y will increase the area; if it is negative, increasing y will decrease the area.

Maximizing Area

Maximizing area involves finding the dimensions that provide the largest possible space within a given constraint. For the Soo family, they needed to maximize the area of a rectangle given a perimeter of 48 ft, with one side against a barn.

This problem is optimized by forming an equation for the area, A(y) = (48 - 2y)y, which relates the dimensions to the fixed perimeter. By differentiating this equation, we find the critical points where the maximum area could occur. The calculations showed that the area would be largest when y = 12 ft, and subsequently, x = 24 ft. This configuration allows them to utilize the space most effectively given the constraints.

Critical Points in Functions

Critical points are values in a function where the derivative is zero or undefined. These points are essential in understanding where the maxima or minima of functions occur. For the Soo family's fence problem, we sought the critical points of the area function to find where the maximum area occurred.

Using the derivative we obtained, A'(y), we found that setting it to zero gave us y = 12. This is our critical point. To determine whether it's a maximum or minimum, we look at the second derivative, A"(y). If A"(y) is negative (which it was in this problem), we have a maximum at that critical point, ensuring that 12 ft was indeed the width for the greatest area given the circumstances.