Math Problem Statement

Une usine d’objets en résine fabrique des boîtiers de portable. La machine fonctionne 7 jours sur 7 durant le mois de juin. La production est de 2 500 boîtiers le 31 mai. À partir du 1er juin, la production augmente de 50 boîtiers par jour. Pour un client, on stocke la production du 11 juin au 24 juin inclus. On nomme Un la production le n-ième jour du mois de juin. 1) Établir la formule donnant Un en fonction de n et calculer la production du 24 juin. 2) Calculer le nombre de boîtiers stockés pour le client. 3) On vend chaque boîtier 1,40 € pièce, prix TTC. Calculer le montant de la facture TTC pour le client.

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.
    • UnU_n: Represents the number of cases produced on the nn-th day of June.

Tasks:

  1. Find the formula for UnU_n as a function of nn and calculate the production for June 24.
  2. Calculate the number of cases stocked for the client between June 11 and June 24 (inclusive).
  3. Determine the total bill for the client if each case costs €1.40 (TTC).

Step-by-Step Solution:

1. Formula for UnU_n:

On May 31, the production was 2500. On June 1, production starts increasing by 50 cases per day.

  • Day n=1n = 1 corresponds to June 1. The production on this day is U1=2500+50=2550U_1 = 2500 + 50 = 2550.
  • For any day nn in June: Un=2500+50nU_n = 2500 + 50n Production on June 24 (n=24n = 24): U24=2500+5024=2500+1200=3700U_{24} = 2500 + 50 \cdot 24 = 2500 + 1200 = 3700

2. Total number of cases for June 11 to June 24 (inclusive):

This requires summing U11U_{11} to U24U_{24}.

  • Formula for the sum of an arithmetic sequence: S=n2(Ufirst+Ulast)S = \frac{n}{2} \cdot (U_{\text{first}} + U_{\text{last}}) Where:
    • nn: Number of terms.
    • Ufirst=U11=2500+5011=3050U_{\text{first}} = U_{11} = 2500 + 50 \cdot 11 = 3050
    • Ulast=U24=3700U_{\text{last}} = U_{24} = 3700
    • n=2411+1=14n = 24 - 11 + 1 = 14

Substitute values: S=142(3050+3700)=76750=47250S = \frac{14}{2} \cdot (3050 + 3700) = 7 \cdot 6750 = 47250

Total cases stocked: 47,250


3. Total bill for the client:

The cost per case is €1.40.

Total bill: Total Cost=472501.40=66150\text{Total Cost} = 47250 \cdot 1.40 = 66150 \, \text{€}


Final Answers:

  1. Formula: Un=2500+50nU_n = 2500 + 50n, U24=3700U_{24} = 3700.
  2. Total stock for the client: 47,250 cases.
  3. Total cost: €66,150.

Let me know if you need further details or assistance!


Questions to expand on:

  1. Can we derive this solution graphically to represent the growth of production?
  2. What would happen if the daily increase changed to 60 cases per day?
  3. How would the total cost change if the price per case was raised by 10%?
  4. What is the average production rate during the month of June?
  5. 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