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A coffee shop currently sells 490 lattes a day at $3.00 each. They recently tried raising the by price by $0.25 a latte, and found that they sold 30 less lattes a day. a) Assume that the number of lattes they sell in a day, N, is linearly related to the sale price, p (in dollars). Find an equation for N as a function of p.

1 Answer

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Answer:

N(p) = -120p + 850

Explanation:

Let the number of lattes sold by the coffee shop be N

and p be the price of each Lattes

The number of lattes (N) sold by a coffee shop is linearly related to the sale price (p).

Now,Let the linear relationship be

N(p) = mp+c, -------------------(1)

where

m is the slope

c is an arbitrary real number.


m = (y_2 -y_1)/(x_2-x_1)--------------------(2)


y_1 = number of lattes sold currently = 490


y_2 = number of lattes sold previously = 490 - 30


x_1 = cost of lattes sold currently = 3.00


x_2 = cost of lattes sold previously = 3.00+0.25

Substituting the values in eq(2)

Now, m =
([(490-30)-490])/([(3.00+0.25)-3.00])

m =
(460-490)/(3.25-3.00)

m =
(-30)/(0.25)

m = -120.

On substituting m = -160 eq(1), we get

N(p) = -120p+c.

Further, on substituting N = 430 and p = 2.75 , we get


490 = -120 * 3.00 +c

490 = -360 + c

490+360 = c

c = 850

Thus

N(p) = -120p + 850

User Bernd Verst
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