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Suppose that an office chair costs $50 in 2017, and $52.90 in 2018. Office chairs continue increasing in price, but this price increase decreases by 10 cents each subsequent year. Use anti-derivatives to express the price of a chair as a function of t, where t represents the number of years that have passed since 2017, so that t= 0 corresponds to the year 2017, t= 1 corresponds to the year 2018, and so on.

User Tomblue
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Final answer:

The price of an office chair as a function of t, given that the price increase decreases by $0.10 each year, can be expressed using an arithmetic sequence within the anti-derivative. The price function is Price(t) = $50 + ½ t [5.70 - 0.10t], where t is the number of years since 2017.

Step-by-step explanation:

The student wishes to express the price of an office chair as a function of t, where t represents the years since 2017. The initial price in 2017 (t=0) is $50. In 2018, it increased by $2.90, and it is given that the rate of increase will decrease by $0.10 each year subsequently. We can model the price change by a function of the form: Price(t) = P0 + ∑i=1t(Ri), where P0 is the initial price, Ri is the rate of change for year i, and ∑ represents the sum.

To find an explicit function for the price, we recognize that the sequence of the rate of change of price forms an arithmetic sequence with a common difference of -$0.10. This leads us to the equation of the arithmetic sequence Rt = R1 - 0.10(t-1). The sum of the first t terms of an arithmetic sequence is given by St = ½ t [2a + (t - 1)d], where a is the first term, and d is the common difference.

The price increase in 2018 (t=1) was $2.90, which makes it the first term in our arithmetic sequence (a = $2.90). The common difference (d) is -$0.10. Substituting these values in the sum equation, we have St = ½ t [2(2.90) + (t - 1)(-0.10)]. The function for the price of the chair in terms of years since 2017 is then Price(t) = $50 + St.

Therefore, Price(t) = $50 + ½ t [5.80 - 0.10(t - 1)] which simplifies to Price(t) = $50 + ½ t (5.70 - 0.10t).

User LTech
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