Problem 13 - Introductory Algebra for College Students

Question

The perimeter of a badminton court is 128 feet. After a game of badminton, a player's coach estimates that the athlete has run a total of 444 feet, which is equivalent to six times the court's length plus nine times its width. What are the dimensions of a standard badminton court?

Step-by-Step Solution

Answer

The dimensions of the badminton court are the length and the width obtained from the solution steps.

Step 1: Forming the First Equation

A badminton court is a rectangle. The perimeter of a rectangle is given by the formula \(2 \times (\text{length} + \text{width})\). From the problem, it's known the perimeter is 128 feet, so the first equation is \(2\times (\text{length}+\text{width}) =128\) which simplifies to \(\text{length}+\text{width} =64\).

Step 2: Forming the Second Equation

According to the coach, the athlete has run 444 feet, which is six times the court's length plus nine times its width. Therefore, the second equation from this information is \(6\times \text{length}+9\times \text{width}=444\).

Step 3: Solving the System of Equations

To solve the system of equations, one could either use substitution or elimination method. Substitution is selected in this case. From equation 1, express width as \(64-\text{length}\). Next substitute \(64-\text{length}\) for width in equation 2, which gives us \(6 \times \text{length}+ 9 \times (64-\text{length})=444\). Solving this equation gives us the length of the court.

Step 4: Find width

After obtaining the length, substitute it into the equation \(\text{length} + \text{width} = 64\) to obtain the width of the court.