Finding the derivative of a revenue function

In summary: And if it did, then r = p * (-1/4 p + 10000) = (-1/4)p^2 + 10000 p, and the derivative dr/dp = (-1/2)p + 10000 at p = 8. If you want the derivative of r = pq, then r = p * (-1/4 p + 10000) = (-1/4)p^2 + 10000 p, and the derivative dr/dp = (-1/2)p + 10000 at p = 8. But that is not what the problem statement asks for. So what is the real problem statement? In summary, the revenue function for a product is r
  • #1
tech_chic
3
0

Homework Statement


The revenue function for a product is r = 8x where r is in dollars and x is the number of units sold. the demand function is q = -1/4p + 10000 where q units can be sold when selling price is p. what is dr/dp?

Homework Equations


r=pq

The Attempt at a Solution


I substituted the values into r=pq.

8x= -1/4p^2 + 10000p

i'm not sure which rule to use from here to find the derivative or if i overcomplicated the question.
 
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  • #2
tech_chic said:

Homework Statement


The revenue function for a product is r = 8x where r is in dollars and x is the number of units sold. the demand function is q = -1/4p + 10000 where q units can be sold when selling price is p. what is dr/dp?

Homework Equations


r=pq

The Attempt at a Solution


I substituted the values into r=pq.

8x= -1/4p^2 + 10000p

i'm not sure which rule to use from here to find the derivative or if i overcomplicated the question.

You got slightly off track.

r = pq and q is a function of p. In essence, you have r = p * f(p). How would you find dr/dp now?
 
  • #3
I'm thinking I would use the product rule. but I'm still confused on the equation 'r = p * f (p).
 
  • #4
tech_chic said:
I'm thinking I would use the product rule. but I'm still confused on the equation 'r = p * f (p).

It's not clear why you are still confused.

The product rule is OK, but is that rule all that you need to find dr/dp?
 
  • #5
oh yeah the chain rule as well. and I'm still partially confused because I'm not sure how to use the rules on the equation.
 
  • #6
Can someone explain why "r = 8x where r is in dollars and x is the number of units sold" and "there are 10000 - p/4 units sold if the price is p" does not lead to r = 8p ?

And hence dr/dp = -2 and revenue is maximum ($ 80000) if you give the product away for free ?

It doesn't sound logical, so either I am misunderstanding, or I am misinformed in the problem formulation
 
  • #7
tech_chic said:
oh yeah the chain rule as well. and I'm still partially confused because I'm not sure how to use the rules on the equation.

Start with applying the chain rule to the equation r = p * f(p). Don't worry about substituting for f(p) at first.
 
  • #8
tech_chic said:

Homework Statement


The revenue function for a product is r = 8x where r is in dollars and x is the number of units sold. the demand function is q = -1/4p + 10000 where q units can be sold when selling price is p. what is dr/dp?

Homework Equations


r=pq

The Attempt at a Solution


I substituted the values into r=pq.

8x= -1/4p^2 + 10000p

i'm not sure which rule to use from here to find the derivative or if i overcomplicated the question.

There is something seriously wrong with your problem statement. If, in fact, the revenue really is r = 8x (x = number sold), then the price must = 8 (=the coefficient of the number sold in the expression for r), and you have x = 10000(8) - (1/4)8^2 = 79984 and r = 8x = 639872. Nothing is varying, so the derivative = 0.

On the other hand, if you want the derivative of r = p*q(p) at p = 8, that is an entirely different question. The equation you wrote (8x = -1/4 p^2 + 10000 p) makes no sense.
 

Related to Finding the derivative of a revenue function

1. What is a revenue function?

A revenue function is a mathematical model that represents the relationship between the quantity of a product or service sold and the corresponding revenue generated. It is typically denoted by R(x) where x represents the quantity sold.

2. Why is it important to find the derivative of a revenue function?

The derivative of a revenue function represents the rate of change of revenue with respect to the quantity sold. This information is crucial for businesses to make decisions about pricing, production, and maximizing profits.

3. How do you find the derivative of a revenue function?

To find the derivative of a revenue function, you can use the power rule, product rule, or quotient rule depending on the form of the function. You can also use the chain rule if the revenue function is composed of multiple functions.

4. Can the derivative of a revenue function be negative?

Yes, the derivative of a revenue function can be negative. This means that as the quantity sold increases, the revenue decreases. This could happen when the product or service has a decreasing demand or if the price decreases.

5. How can I use the derivative of a revenue function to maximize profits?

To maximize profits, you can set the derivative of the revenue function equal to zero and solve for the quantity that will yield the maximum revenue. This is known as the critical point. Additionally, you can use the second derivative test to determine if the critical point is a maximum or minimum point.

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