Math Problem Statement
Solution
Let’s analyze and solve the problem step by step.
Problem Overview:
- Production Information:
- On May 31: 2500 cases are produced.
- From June 1 onwards: Production increases by 50 cases daily.
- : Represents the number of cases produced on the -th day of June.
Tasks:
- Find the formula for as a function of and calculate the production for June 24.
- Calculate the number of cases stocked for the client between June 11 and June 24 (inclusive).
- Determine the total bill for the client if each case costs €1.40 (TTC).
Step-by-Step Solution:
1. Formula for :
On May 31, the production was 2500. On June 1, production starts increasing by 50 cases per day.
- Day corresponds to June 1. The production on this day is .
- For any day in June: Production on June 24 ():
2. Total number of cases for June 11 to June 24 (inclusive):
This requires summing to .
- Formula for the sum of an arithmetic sequence:
Where:
- : Number of terms.
Substitute values:
Total cases stocked: 47,250
3. Total bill for the client:
The cost per case is €1.40.
Total bill:
Final Answers:
- Formula: , .
- Total stock for the client: 47,250 cases.
- Total cost: €66,150.
Let me know if you need further details or assistance!
Questions to expand on:
- Can we derive this solution graphically to represent the growth of production?
- What would happen if the daily increase changed to 60 cases per day?
- How would the total cost change if the price per case was raised by 10%?
- What is the average production rate during the month of June?
- Can we create a generalized formula for similar problems with variable initial production and daily increase?
Tip: Always double-check arithmetic calculations for large sums in real-world scenarios like invoices!
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Math Problem Analysis
Mathematical Concepts
Arithmetic Sequences
Algebra
Summation of Sequences
Formulas
Un = U0 + d × n
Sum of an arithmetic sequence: S = n/2 × (first term + last term)
Total cost = number of items × price per item
Theorems
Arithmetic Sequence Formula
Suitable Grade Level
Grades 10-12
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