|
Question 466150: (Revenue) The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. If the revenue R, in dollars, is
R(p)=-1/2p²+1900p
• (a) At what prices p is revenue zero?
• (b) For what range of prices will revenue exceed $1,200,000?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price p, in dollars, that it charges. If the revenue R, in dollars, is
R(p)=-1/2p²+1900p
• (a) At what prices p is revenue zero?
---
Solve: (-1/2)p^2+1900p = 0
---
Factor:
p[(-1/2)p + 1900] = 0
---
p = 0 or p = 3800
==============================
• (b) For what range of prices will revenue exceed $1,200,000?
Solve: (-1/2)p^2+1900p > 1,200,000
------------
(-1/2)p^2+1900p - 1200000 > 0
===
(-(1/2))[p^2-3800p+ 2400000 > 0
---
p^2 - 3800p + 2400000 < 0
----------------------------
Graph:
===========================
Solution: 800 < p < 3000
---
Cheers,
Stan H.
|
|
|
| |
