Revenue Question with Linear Function HELP

In summary, if you want to sell the Pen Tablets for the largest possible monthly revenue, you should sell them for $140.
  • #1
AnnaSuxCalc
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0
Revenue Question with Linear Function **HELP**

Homework Statement


(a) Express revenue (R=pq) as a function solely of p
(b) Sketch the graph of R as a function of p. Clearly label the x-intercept(s), y-intercept(s), and vertex (if there is one)
(c) At what price should you sell the Pen Tablets to get the largest revenue? What is the largest possible monthly revenue?

Homework Equations


q = -0.5p + 140 <-- demand function for Pen Tablets
q --> number of Pen Tablets that can be sold/month
p --> price ($) per Pen Tablet


The Attempt at a Solution


Once again I need a little help getting started on part (a), how can I make this be a function of only p
:eek:
PLEASE HELP get my STARTED on this!
 
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  • #2


a) If R = p*q and q = -0.5p + 140, can you rewrite R solely in terms of p?
 
  • #3


Mark44 said:
a) If R = p*q and q = -0.5p + 140, can you rewrite R solely in terms of p?

Mark:!) Yes I can lol:
R = p(-0.5p + 140)
R = -0.5p2 + 140p
 
  • #4


Well, good!
What does the graph look like (part b)? Be sure to label the vertex, since that figures into part c.
 
  • #5


R = -0.5p2 + 140p --> y = -0.5x2 + 140x
Use the Quadratic Formula to find the two values of x:
x=-140 +/- [tex]\sqrt{140^2 - 4(-0.5)(0)}[/tex] /2(-0.5)
x = 280 OR 0

when x = 280
y = -0.5(280)2 + 140(280)
y = -39200 + 39200
y = 0
when x = 0
y = -0.5(0)2 + 140(0)
y = 0

--> therefore the two intercepts are (280,0) and (0,0)
Vertex:
h=-b/2a
h=-140/2(-0.5)
h=140

y=-0.5(140)2 + 140(140)
y=9800

--> therefore the vertex is (140,9800)

now, for part (c)which one is which:blushing:
 
  • #6


Well how do you find the largest anything in calculus?
 
  • #7


NoMoreExams said:
Well how do you find the largest anything in calculus?

Uuhm...I have no Idea:confused:
dont I use the vertex to find this?
so --> the vertex is (140,9800)
is the 140 the p ? cos if it is I would sub p into the equation q = -0.5p + 140
so that q = -0.5(140) + 140
then q = 70
THEN
R = pq = 70*140 = 9800 ---> lol you I knew the R was the y part of the vertex...
anyways if NOT I just did a whole lot of typing for nothing :rolleyes:
 
  • #8


Why did you change to x and y? Now you don't know which one is which. The variable p (unit price) and R (total revenue) would be more helpful to keep things straight.

And here's the question for c):
At what price should you sell the Pen Tablets to get the largest revenue? What is the largest possible monthly revenue?
 
  • #9


I know Mark lol...I have no Idea why I changed it..it seemed easier at the time:redface:
sooo the fact that your not responding to my calculation means that they are...wrong?!
I drew the graph and the y-axis is the R and the x-axis is the p
so to answer (c):
I should sell the Pen Tablets at $140 to get the largest revenue AND The largest possible monthly revenue is $9800!
Correct?!
 
  • #10


Looks good.
 
  • #11


Woohoo!:!)
 

FAQ: Revenue Question with Linear Function HELP

What is a linear function?

A linear function is a mathematical equation that represents a straight line on a graph. It has the form f(x) = mx + b, where m is the slope of the line and b is the y-intercept. It is used to model relationships between two variables that have a constant rate of change.

How is a linear function used to calculate revenue?

In business, a linear function is often used to model revenue. The slope of the line represents the rate of change of revenue over time, and the y-intercept represents the initial revenue. By plugging in different values for x (time), you can calculate the revenue at different points in time.

What is the revenue question with linear function?

The revenue question with linear function is a specific application of a linear function where the independent variable (x) represents time and the dependent variable (y) represents revenue. The equation is typically of the form y = mx + b, where m is the rate of change of revenue over time and b is the initial revenue.

How can I use a linear function to predict future revenue?

Once you have a linear function that models your revenue, you can use it to predict future revenue by plugging in a value for x that represents a future point in time. The resulting y-value will be the predicted revenue at that time.

What are some limitations of using a linear function to model revenue?

Linear functions assume that the relationship between time and revenue is constant, which may not always be the case in real-world scenarios. Additionally, they do not account for external factors that may impact revenue, such as changes in the market or consumer behavior. It is important to regularly review and adjust the linear function to ensure it accurately reflects the revenue trend.

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