applications of integration: revenue

[calcissoannoyingugh]

I'm stuck on this problem because I'm not sure if you need to integrate the equation to get the answer or not. (I integrated because the homework covers the topic applications of integration but it just feels like it doesn't make sense to.)

A girl sells lemonade. The price per ounce ? (in pennies) for which she sells her lemonade is given by ? = 8 − 0.001(? − 50) 2 where ? is the number of ounces of lemonade that she has already sold. Revenue is price times quantity, i.e. ? = ??. Find the revenue earned by selling a gallon of lemonade. One gallon equals 128 ounces.

the equation for price simplifies to -0.001x^2 + 0.1x + 5.5 and after integrating I get an equation of -0.00033x^3 + 0.05x^2 + 5.5x and evaluating over the interval from 0 to 128 gives me an answer of $831.14
would that be correct or do i just need to plug in 128 into the original equation to get the right answer?

 

[Harry_the_cat]

You are asked to find revenue R when x = 128.
The formula for revenue is given ie R=Px = [8-0.001(x-50)^2] * x.
Just sub in to the formula for R. There is nothing to integrate.

 

[lookagain]

Revenue is price times quantity, i.e. ? = ??.

R = revenue

R = px \(\displaystyle \ \ \ \ \ \ \ \ \) The p is lower case for the price.

P = profit

C = cost

P = R - C