Problem 25 - Algebra 2

Question

construction A construction company will be fined for each day it is late completing a bridge. The daily fine will be \(\$ 4000\) for the first day and will increase by \(\$ 1000\) each day. Based on its budget, the company can only afford \(\$ 60,000\) in total fines. What is the maximum number of days it can be late?

Step-by-Step Solution

Answer

The maximum number of days the company can be late without exceeding the budget is 8 days.

Step 1: Understand the Problem

The company faces a fine that starts at $4000 for the first late day and increases by $1000 for each subsequent day. They can afford up to $60,000 in fines. We need to find the maximum days they can be late without exceeding this budget.

Step 2: Define the Arithmetic Sequence

Recognize that the fines form an arithmetic sequence, where the first term is $4000 and the common difference is $1000. The sequence is: $4000, $5000, $6000, ... for each late day.

Step 3: Formula for Sum of Arithmetic Sequence

Use the formula for the sum of an arithmetic sequence: \[ S_n = \frac{n}{2} \times (a_1 + a_n) \]where \( S_n \) is the sum of the fines (less than or equal to $60,000), \( a_1 = 4000 \), \( a_n \) is the amount on the nth day, and \( n \) is the number of days late.

Step 4: Express Conditions in the Formula

Set up the equation with the known values:\[ 60000 = \frac{n}{2} \times (4000 + (4000 + (n-1) \times 1000)) \]Simplifying, we have:\[ 60000 = \frac{n}{2} \times (8000 + 1000(n-1)) \]

Step 5: Solve for the Maximum Number of Days (n)

Continue simplifying and solving for \( n \):\[ 60000 = \frac{n}{2} \times (8000 + 1000n - 1000) \]\[ 60000 = \frac{n}{2} \times (1000n + 7000) \]\[ 120000 = n \times (1000n + 7000) \]\[ 120000 = 1000n^2 + 7000n \]\[ 0 = 1000n^2 + 7000n - 120000 \]Now solve using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1000 \), \( b = 7000 \), \( c = -120000 \).

Step 6: Calculate the Quadratic Formula

Calculate discriminant:\[ b^2 - 4ac = 7000^2 - 4 \times 1000 \times (-120000) \]\[ = 49000000 + 480000000 = 529000000 \]Now find \( n \):\[ n = \frac{-7000 \pm \sqrt{529000000}}{2000} \]\[ n = \frac{-7000 \pm 23000}{2000} \] Select the positive solution:\[ n = \frac{16000}{2000} = 8 \].

Step 7: Verify the Solution

Ensure the sum of fines for 8 days doesn't exceed \(60,000:The fines for 8 days are: \)4000, \(5000, \)6000, ..., $11000.Summing these gives:\[ 4,000 + 5,000 + 6,000 + 7,000 + 8,000 + 9,000 + 10,000 + 11,000 = 60,000 \] This sum equals the budget limit, confirming the solution is correct.

Key Concepts

quadratic formula

The quadratic formula is a powerful tool used to solve quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). Here, the coefficients \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows us to find the values of \( x \) that make the quadratic equation true. In the context of our exercise, the quadratic formula helps determine the maximum number of days the construction company can afford to delay its project without exceeding the fine budget of \$60,000.
To use the quadratic formula, always follow these steps:

• Identify the coefficients \( a \), \( b \), and \( c \) from your quadratic equation.
• Calculate the discriminant \( b^2 - 4ac \) to check if the solutions are real and distinct.
• Apply the formula to solve for the variable \( x \).

In our exercise, the solution is clear and shows that the quadratic formula gives \( n = 8 \), verifying the company can be late up to 8 days.

arithmetic sequence formula

An arithmetic sequence is a set of numbers in which the difference between any two consecutive terms is constant. This constant difference is known as the common difference. The arithmetic sequence formula allows us to express any term in the sequence based on its position. The formula is:\[a_n = a_1 + (n-1) \cdot d\]where:

• \( a_n \) is the \( n^{th} \) term.
• \( a_1 \) is the first term.
• \( n \) is the term number.
• \( d \) is the common difference.

In the construction company's problem, the fines form an arithmetic sequence with \( a_1 = 4000 \) and \( d = 1000 \). Each day the fine increases by \$1000, forming a sequence like 4000, 5000, 6000, etc. This helps in setting up our problem so we can calculate the total sum of fines over the days.

maximum number of days

To determine the maximum number of days the construction company can afford to be late, it's crucial to ensure the total fines do not surpass \$60,000. The fines are structured as an arithmetic sequence, and their sum must remain within budget constraints.
To find this maximum number of days \( n \), we relate the sum of the arithmetic sequence, represented by:\[S_n = \frac{n}{2} \times (a_1 + a_n)\]We substitute in to set an equation:\[60000 = \frac{n}{2} \times (4000 + (4000 + (n-1) \cdot 1000))\]After solving the equation using the quadratic formula, we determine that the maximum \( n \) is 8. This calculation confirms that the company can afford to be late up to 8 days while staying within its budget.

sum of sequence

Calculating the sum of an arithmetic sequence is essential in understanding how multiple terms add up to a total value. The sum formula is central to problems like the one faced by the construction company, where multiple incremented fines contribute to a total late fee.
The formula for the sum of the first \( n \) terms of an arithmetic sequence is:\[S_n = \frac{n}{2} \times (a_1 + a_n)\]Where:

• \( S_n \) is the sum of the first \( n \) terms.
• \( a_1 \) is the first term, \( 4000 \) in this problem.
• \( a_n \) is the \( n^{th} \) term or \( a_1 + (n-1)d \).
• \( n \) is the number of terms.

For the company's fine sequence, this formula ensures that when we compute the sum for \( n = 8 \), the total fine equals the maximum allowed \$60,000. This confirmatory step is vital in verifying the solution's accuracy and demonstrating the application of arithmetic series in real-world situations.