1. ## Maximize Revenue

Word problem, stuck at part a. (doh) I know that R=price*quantity, but i'm not sure how to set this one up and what to get the derivative of.

demand function for corn:
p=6,570,000/q^1.3

q is the number of bushels of corn that could be sold at p dollars per bushel in one year. Assume that at least 10,000 bushels must be sold each year.

a) How much should farmers charge per bushel to maximize annual revenue?

b) how much corn can farmers sell each year at that price?

c) what is the farmers' resulting revenue?

2. so, if i write the equation R= (6,570,000/q^1.3)(q), then that gives me R in terms of q, when i need to maximize r in terms of p? i'm just confused silly here, and would appreciate a nudge in the right direction(or a violent shove)

3. maximum revenue occurs when marginal revenue is 0 (i.e. the derivative of the revenue function is 0). Take the derivative of R and set it equal to 0 and find which value of q will satisfy that. From this you will get the price.

4. aha, thanks

oh, it is a bit of a hairy one, not sure what the derivative would be, do i use the product rule?

5. $R = 6570000q^{-0.3}$ so you can just use the power rule.

$-1971000q^{-1.3}$ is the derivative.

6. isnt it 6570000/q^.3 or 6570000q^-.3? hm

7. Thats right I screwed up. Fixed it.

8. so this confuses me more, heh, how is it that revenue can be such a large negative number?
and q cannot be any less than 10,000, so it's not like i can make q a negative number to make revenue positive.

Hmmm...

9. Originally Posted by leviathanwave
so this confuses me more, heh, how is it that revenue can be such a large negative number?
and q cannot be any less than 10,000, so it's not like i can make q a negative number to make revenue positive.

Hmmm...
the revenue is not negative. it's derivative is, there's a difference.

by the way, are you sure your demand function is correct? i can't get zero for either derivative (i solved for R in terms of p, which is what you should have done. the derivative here is never zero)