Math Problem Statement

Solve question

A small trading company made a profit of $250,000 in the year 2000. The company considered two different plans, Plan A and Plan B, for increasing its profits. Under Plan A, the annual profit would increase each year by 5% of its value in the preceding year. Find for Plan A, (i) the profit for the year 2008, (ii) the total profit for the 10 years from 2000 to 2009 inclusive. Under Plan B, the annual profit would increase each year by a constant amount D. (iii) Find the value of D for which the total profit for the 10 years from 2000 to 2009 inclusive would be the same for both plans.

Solution

Let's break down the problem and solve it step by step:

Given:

  • The company made a profit of $250,000 in the year 2000.
  • Two plans for increasing profits are considered:
    • Plan A: The annual profit increases each year by 5% of its value in the previous year.
    • Plan B: The annual profit increases by a constant amount $D each year.

We are tasked with:

  1. For Plan A:

    • Find the profit for the year 2008.
    • Find the total profit for the 10 years (from 2000 to 2009 inclusive).
  2. For Plan B:

    • Find the value of $D such that the total profit for the 10 years under Plan B is the same as the total profit for the 10 years under Plan A.

Part 1: Plan A

(i) Finding the profit for the year 2008 under Plan A:

The profit grows by 5% each year, so this is a geometric sequence.

  • The initial profit in 2000 is $250,000.
  • The formula for the profit in year nn (where nn starts at 0 for 2000) is: Pn=250,000×(1.05)nP_n = 250,000 \times (1.05)^n

For the year 2008, n=8n = 8 (since 2008 is 8 years after 2000): P8=250,000×(1.05)8P_8 = 250,000 \times (1.05)^8

Let's calculate that.

(ii) Finding the total profit for the 10 years (2000 to 2009 inclusive) under Plan A:

The total profit for the 10 years is the sum of a geometric series: S10=250,000×(1.05101)0.05S_{10} = 250,000 \times \frac{(1.05^{10} - 1)}{0.05} This gives the total profit from 2000 to 2009.

Part 2: Plan B

(iii) Find the value of $D such that the total profit under Plan B is the same as under Plan A.

Under Plan B, the profit increases by a constant amount $D each year, which forms an arithmetic sequence.

  • The profit in 2000 is $250,000.
  • The formula for the profit in year nn under Plan B is: Pn=250,000+n×DP_n = 250,000 + n \times D

The total profit for the 10 years (2000 to 2009) under Plan B is the sum of an arithmetic series: S10=102×(2×250,000+9×D)S_{10} = \frac{10}{2} \times (2 \times 250,000 + 9 \times D) We can equate this to the total profit for Plan A and solve for $D.


I'll start by calculating the results for Plan A first.### Plan A:

  1. The profit for the year 2008 is $369,364 (rounded to 3 significant figures).
  2. The total profit for the 10 years (2000 to 2009 inclusive) is $3,144,473 (rounded to 3 significant figures).

Next, I'll solve for the constant increase $D in Plan B by equating the total profit for the 10 years under both plans. Let's proceed with that calculation.### Plan B:

The constant increase $D for which the total profit over the 10 years (2000 to 2009) would be the same as in Plan A is approximately $14,322 (rounded to 3 significant figures).

Would you like further clarifications or more details? Here are some related questions:

  1. How would the profits in 2005 compare under both plans?
  2. How can we adjust the growth rate in Plan A to match a specific target?
  3. What if the initial profit was higher in 2000? How would that impact the results?
  4. How would inflation adjustments affect the total profits in either plan?
  5. Can we compute the break-even point between Plan A and Plan B over a different period?

Tip: For geometric sequences, each term is multiplied by a constant factor, whereas in arithmetic sequences, a constant amount is added to each term. Understanding this difference is key to solving similar problems.

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Math Problem Analysis

Mathematical Concepts

Geometric sequences
Arithmetic sequences
Profit growth
Series summation

Formulas

P_n = 250,000 * (1.05)^n (for geometric sequence growth in Plan A)
S_10 = 250,000 * ((1.05^10 - 1)/0.05) (total profit in Plan A)
P_n = 250,000 + n * D (for arithmetic sequence growth in Plan B)
S_10 = (10/2) * (2 * 250,000 + 9 * D) (total profit in Plan B)

Theorems

Sum of geometric series
Sum of arithmetic series

Suitable Grade Level

Grades 10-12