|
Question 167650: I really do not understand this problem at all... help, please! ---use the fact that revenue = price*quantity to solve the problem. the price (p) in dollars and the quantity (x) sold of a certain product are described by the following: p=-1/6x+100, 0<=x<=600.
a)express the revenue R as a function of x.
b)what quantity x maximizes the revenue?
c)what price should the company charge for the maximum revenue?
Found 2 solutions by stanbon, gonzo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! use the fact that revenue = price*quantity to solve the problem. the price (p) in dollars and the quantity (x) sold of a certain product are described by the following: p=-1/6x+100, 0<=x<=600.
a)express the revenue R as a function of x.
R(x) = x[(-1/6)x + 100] = (-1/6)x^2 + 100x
-------------------------------------
b)what quantity x maximizes the revenue?
max occurs when x = -b/2a = -100/(2(-1/6)) = 300
--------------------------------------
c)what price should the company charge for the maximum revenue?
P(300) = (-1/6)(300) + 100
p(300) = -50 + 100 = $50
========================================
Cheers,
Stan H.
Answer by gonzo(654) (Show Source):
You can put this solution on YOUR website! let r = revenue
let p = price
let x = quantity
----
r = p*x
p = (-1/6)*x + 100
-----
substituting the formula for p in the revenue equation, we get:
r = ((-1/6)*x + 100)*x
multiplying out to remove parentheses, this becomes:
r = (-1/6)*x^2 + 100*x
this formula expresses revenue as a function of x.
-----
second part says what quantity of x maximizes the revenue?
-----
easiest way to solve this unless you have to do it algebraically, is to graph the equation and see where the maximum lies.
-----
graph of the equation looks like this:
look below the graph for further comments.
-----
the graph shows that the maximum revenue will occur when x is around 300 and that the maximum revenue will be around 15000.
-----
since the graph is a quadratic, we can use some of the formulas for quadratic equations to find out where the maximum value of the formula lies exactly.
-----
the formula for max/min of a quadratic equation is -b/2a.
in the formula -(1/6)*x^2 + 100*x,
a = (-1/6), and
b = 100
-----
-b/2a = -100/(-1/3)
this is the same as:
-100*(-3) which equals 300.
-----
the maximum / minimum value of the equation occurs at x = 300.
since the a value is negative, the equation opens downward and points upward, so the value of x = 300 is a maximum value.
-----
when x = 300, the value of r = (-1/6)*(300)^2 + 100*300.
this becomes:
(-1/6)*90000 + 30000
this becomes:
-15000 + 30000
this becomes:
15000
-----
price = (-1/6)*300 + 100 = $50.00 per goods sold.
-----
maximum revenue = $15,000 when 300 goods are sold at $50.00 apiece.
-----
|
|
|
| |
