First week sells 6 chairs for $80 each. Next week, if chair is not sold, it will sell for 0.85 times previous week's price. Store needs to sell 6 chairs for total of $270 to make a profit. what is the last week in which all 6 chairs could be sold so that the store makes a profit?

Question

First week sells 6 chairs for $80 each. Next week, if chair is not sold, it will sell for 0.85 times previous week's price. Store needs to sell 6 chairs for total of $270 to make a profit. what is the last week in which all 6 chairs could be sold so that the store makes a profit?

Answer 1

The worst case is when all six chairs are sold at the lowest acceptable price, which is 270/6 = $45 each.

So, you want w where

80*0.85^w = 45
w = 3.54

So, the 3rd week is the last possible week to make a profit.

80*.85^3 = $49.13
80*.85^4 = $41.76 - too low

Answer 2

To find the last week in which all 6 chairs could be sold so that the store makes a profit, we can work backwards.

Let's start with the last week (week 4) and the given condition that the total sum should be $270.

Let's assume the selling price for the chairs in week 4 is x. Since we know that in the next week the price is reduced to 0.85 times the previous week's price, the selling price in week 3 will be 0.85x, in week 2 it will be (0.85)^2 * x, and in week 1 it will be (0.85)^3 * x.

Since the store needs to sell 6 chairs each week, the total sum of the selling prices for each week should be $270. Therefore, we can set up the following equation:

x + 0.85x + (0.85)^2 * x + (0.85)^3 * x = $270

Simplifying the equation:

x (1 + 0.85 + (0.85)^2 + (0.85)^3) = $270

Calculating the sum in the parentheses:

(1 + 0.85 + (0.85)^2 + (0.85)^3) = 3.6965

Substituting back into the equation:

3.6965 * x = $270

Dividing both sides by 3.6965 to solve for x:

x = $270 / 3.6965 ≈ $73.01

So, the selling price in the last week should be approximately $73.01 for the store to make a profit.