| 
October 16th, 2007, 04:40 PM
| | Junior Member | | Join Date: Jan 2007
Posts: 35
Country:  Thanks: 37
Thanked 0 Times in 0 Posts
| |  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? |
October 16th, 2007, 06:44 PM
| | Junior Member | | Join Date: Jan 2007
Posts: 35
Country: Thanks: 37
Thanked 0 Times in 0 Posts
| |
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  ) |
October 16th, 2007, 06:50 PM
|  | Senior Member | | Join Date: Mar 2007
Posts: 307
Country: Thanks: 49
Thanked 106 Times in 95 Posts
| |
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. | | The following users thank tukeywilliams for this useful post: | |  |
October 16th, 2007, 06:58 PM
| | Junior Member | | Join Date: Jan 2007
Posts: 35
Country: Thanks: 37
Thanked 0 Times in 0 Posts
| |
aha, thanks
oh, it is a bit of a hairy one, not sure what the derivative would be, do i use the product rule? |
October 16th, 2007, 07:02 PM
| | Senior Member | | Join Date: Mar 2007
Posts: 307
Country: Thanks: 49
Thanked 106 Times in 95 Posts
| |
 so you can just use the power rule.  is the derivative. | | The following users thank tukeywilliams for this useful post: | | |
October 16th, 2007, 07:04 PM
| | Junior Member | | Join Date: Jan 2007
Posts: 35
Country: Thanks: 37
Thanked 0 Times in 0 Posts
| |
isnt it 6570000/q^.3 or 6570000q^-.3? hm |
October 16th, 2007, 07:06 PM
| | Senior Member | | Join Date: Mar 2007
Posts: 307
Country: Thanks: 49
Thanked 106 Times in 95 Posts
| |
Thats right I screwed up. Fixed it. | | The following users thank tukeywilliams for this useful post: | | |
October 16th, 2007, 07:08 PM
| | Junior Member | | Join Date: Jan 2007
Posts: 35
Country: Thanks: 37
Thanked 0 Times in 0 Posts
| |
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...
Last edited by leviathanwave; October 16th, 2007 at 07:30 PM.
|
October 16th, 2007, 07:54 PM
|  | vs Jhevon | | Join Date: Feb 2007 Location: New York, USA
Posts: 11,590
Country:  Thanks: 2,668
Thanked 4,470 Times in 4,155 Posts
| |
Quote:
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) | | The Following 2 Users Say Thank You to Jhevon For This Useful Post: | | | | Thread Tools | | | | Display Modes |  Linear Mode | 
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts
HTML code is Off | | | All times are GMT -7. The time now is 04:23 PM. | | |
