A gas station sells 1500 gallons of gasoline per hour if it

A gas station sells 1500 gallons of gasoline per hour if it charges $ 2.15 per gallon but only 1200 gallons per hour if it charges $ 2.60 per gallon. Assuming a linear model

(a) How many gallons would be sold per hour of the price is $ 2.20 per gallon?
Answer:

(b) What must the gasoline price be in order to sell 1400 gallons per hour?
Answer: $

(c) Compute the revenue taken at the four prices mentioned in this problem -- $ 2.15, $ 2.20, $ 2.60 and your answer to part (b). Which price gives the most revenue?
Answer: $

(d) What is the price that the gas station should charge to maximize revenue?
Answer: $

Solution

since we have assumed it as an linear model then we need to set

an independent variable and a dependent variable.

here we have got two constraints the gallons of gasoline sold and the cost of gasoline per hour

let the gallons of gasoline be = y

and the cost per gallons be = x

so initially we have (2.15 , 1500) this we could take as the reference point.

and later on another point on the linear model is (2.6 , 1200)

now we\'ll find the slope m = (y2-y1)/(x2-x1) = (1200 - 1500) /(2.6 - 2.15)

                                  m = -300/ .45 = -100/.15 = -20/.3 = -200/3

point slpope form (y-y1) = m(x-x1)

                          (y-1500) = -200/3(x-2.15)

y= -200x/3 + 1643.33 ----------> this is the linear model

a> when x = 2.20

y = -200(2.20)/3 + 1643.33

y = 1496.66 gallons

b> when y = 1400

1400 = -200x/3 + 1643.33

x = $ 3.65 per gallon

c> when x= 2.15

using y= -200x/3 + 1643.33

y = 1500

total revenue = 1500 * 2.15 = $ 3225

when x = 2.20

y = 1496.66

total revenue = 1496.66 * 2.20 = $ 3292.65

when x = 2.6

y= 1470

total revenue = 1470*2.6 = $ 3822

and from part (b) x = $ 3.65

y = 1400

total revenue = 1400*3.65 = $ 5110

x = $ 3.65 per gallon gives the most price

d> to maximize the revenue the gas station should sharge = $ 3.65 per gallon