Question

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.

Answer 1

Answer:

N(p) = -120p + 850

Step-by-step 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.

$m = \frac{y_2 -y_1}{x_2-x_1}$--------------------(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 =$\frac{[(490-30)-490]}{[(3.00+0.25)-3.00]}$

m = $\frac{460-490}{3.25-3.00}$

m = $\frac{-30}{0.25}$

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

$490 = -120 \times 3.00 +c$

490 = -360 + c

490+360 = c

c = 850

Thus

N(p) = -120p + 850