Problem 13 - Beginning and Intermediate Algebra

Question

Solve. Every winter Rich makes a rectangular ice rink in his backyard. He has \(100 \mathrm{ft}\) of material to use as the border. What is the maximum area of the ice rink?

Step-by-Step Solution

Answer

The maximum possible area of the ice rink is \(625\mathrm{ft}^2\) when the length and the width are both \(25\mathrm{ft}\).

Step 1: Define the variables and equations for the problem

Let the length of the ice rink be \(x \mathrm{ft}\) and the width be \(y \mathrm{ft}\). The perimeter of the rectangle is given by: \(P = 2x + 2y\) Since Rich has 100 ft of material, \(2x + 2y = 100\) The area of the rectangular ice rink can be calculated as: \(A = x*y\) Our goal is to maximize the area, A, with the constraint \(2x + 2y = 100\).

Step 2: Solve the constraint equation for y

Solve the perimeter equation for y: \(2y = 100 - 2x\) \(y = 50 - x\)

Step 3: Substitute the expression for y in the area equation

Now substitute the expression for y in the area equation: \(A = x * (50 - x)\)

Step 4: Find the dimensions that maximize the area by taking the derivative

To find the maximum area, we need to find the dimensions where the first derivative of the area with respect to x is zero. Find the derivative of A with respect to x: \(A'(x) = \frac{d}{dx} (x * (50 - x))\) Apply product rule: \(A'(x) = (50 - x) - x\) Set the derivative equal to zero: \(0 = (50 - x) - x\) Solve for x: \(2x = 50\) \(x = 25\mathrm{ft}\)

Step 5: Calculate the width y using the constraint equation

Now find the width y using the constraint equation: \(y = 50 - x\) \(y = 50 - 25\) \(y = 25\mathrm{ft}\)

Step 6: Find the maximum area using the dimensions

Finally, calculate the maximum area using the dimensions obtained: \(A = x * y\) \(A = 25\mathrm{ft} * 25\mathrm{ft}\) \(A = 625\mathrm{ft}^2\) The maximum possible area of the ice rink is 625 square feet.

Key Concepts

Perimeter Constraint

A perimeter constraint is a restriction placed on the total length of the boundary of a shape, like a rectangle, in a problem. In this context, when designing a rectangular ice rink, the total material Rich has available is 100 feet. This becomes the fixed perimeter of the rink. The formula to calculate the perimeter of a rectangle is given by:

• \( P = 2x + 2y \)

where \(x\) is the length and \(y\) is the width of the rectangle. Since the available border material measures 100 feet, we set:

• \( 2x + 2y = 100 \)

This equation is essential in the optimization problem, as it limits and defines the possible dimensions of the rectangle, ensuring that the ice rink's design remains within Rich's resources.

Area Maximization

The next logical step is to determine how to maximize the area of the rectangular ice rink within the given perimeter condition. The area, often desired to be as large as possible for practical benefits like more skating space, can be defined by the formula:

• \( A = x \times y \)

Here we aim to find the largest possible value for \(A\), the area, by manipulating the dimensions \(x\) and \(y\), while respecting the perimeter constraint of 100 feet.
Utilizing the relationship derived from the perimeter constraint, such as \( y = 50 - x \), allows us to substitute the expression for \(y\) into the area equation, turning it into a single-variable problem. Hence the area equation becomes:

• \( A = x \times (50 - x) \)

This transformation is central for employing calculus methods to find the optimal dimensions that give the maximum possible area.

Derivative Application

Applying calculus, specifically derivatives, allows us to find maximums or minimums of functions—in this case, the area function. To maximize the area, we take the first derivative of the area function with respect to the variable \(x\):

• \( A'(x) = \frac{d}{dx}(x \times (50 - x)) \)

Using the product rule, this becomes:

• \( A'(x) = (50 - x) - x \)

Setting the derivative equal to zero will find the critical points.
Hence,

• \( 0 = 50 - 2x \)

Solving this gives us \( x = 25 \). The derivatives help identify the point where the area is neither increasing nor decreasing, which corresponds to the maximum area in this scenario.

Rectangular Geometry

Understanding rectangular geometry significantly aids in visualizing and solving this type of optimization problem. A rectangle is a four-sided polygon characterized by sides arranged at right angles, with opposite sides being equal. In the ice rink problem, knowing that a perfect square (where length equals width) yields the largest area for a given perimeter can guide reasoning without full calculations.
Since both \(x\) and \(y\) were solved to be 25 feet, the ice rink forms a square.

• This follows from a basic principle in geometry where for a given perimeter, the square encloses the maximal possible area.

The calculated dimensions

• \(25 \times 25\)

yield a maximum area of 625 square feet. This demonstrates the balance between perimeter constraints and optimal area configuration within rectangular geometry.