A coffeeshop determines that it can sell 3000 cups of coffee at a price of $ 1.50

Question

A coffeeshop determines that it can sell  cups of coffee at a price of $, but for every $1 increase in the price, their sales will decrease by 4500 cups and for every $1 decrease, their sales will increase by 4500 cups.

A function giving the revenue earned when the price of coffee is dollars per cup is

The coffeeshop can earn a maximum of_______  when the price of a cup of coffee is____

I do not know what the equation will be for R(x) because there are many variables to consider. Also how would I maximize the amount of coffee? set the equation for R(x)=0??? please help!!!

Answer 1

Bri

If the price is x per cup, let's work out the sales in cups first.

We know when x=1.50 , sales = 3000

And if we decrease x by 1.00 sales will go up by 4500. So if x = 0.5 sales = 3000 + 4500 = 7500

So if y are the sales when x = 1.50 y = 3000 when x = 0.50 y = 7500

x1 = 1.50 y1 = 3000 x2 = 0.50 y2 = 7500

y = mx + c m = (y2-y1)/(x2-x1) = (7500-3000) / (0.5 - 1.5) = -4500

So 3000 = 1.50 x - 4500 + C, so C = 9750

So if x is the price per cup, sales = -4500x + 9750

Revenue = no of cups x price = x (-4500x + 9750) = -4500x² + 9750x

Lets call this R.

To find the value of x that maximises R, solve dR/dx = 0.

Will leave that as an exercise for you !

Mike