Demand Equation x = -20p + 500, 0 <= p <= 25; express R as

[lilcherbear]

If you are given a demand equation x=-20p+500 where 0 is less then or equal to p which is less than or equal to 25, how can you express the revenue R as a function of x?

 

[lilcherbear]

Re: Demand Equation

If I am expressing the revenue R as a function of x for the above problem, would it be -1/20x^2+25x where 0 less than or equal to x which is less than or equal to 500?

 

[stapel]

What is the definition of the variable "p"? How does "demand" relate to "revenue"?

Thank you!

Eliz.

 

[lilcherbear]

Hello,

I understand how revenue and demand are related. What I am not sure is how to create a function representing this based on the above formula. I get the following:

-1/20x^2+25x where 0 less than or equal to x which is less than or equal to 500

Is this correct? Please help.

Thank you.
Cherie

 

[stapel]

I understand how revenue and demand are related. What I am not sure is how to create a function representing this based on the above formula.

You've been given a "demand" function, and you understand the relationship between "demand" and "revenue" (namely, that revenue is the income from selling x units at some price p). So... I'm not understanding what the difficulty is...? Plug the demand function into the relationship between demand and revenue, and thus create the revenue function. :wink:

-1/20x^2+25x where 0 less than or equal to x which is less than or equal to 500

Please reply showing your work and reasoning. Thank you!

Eliz.

 

[lilcherbear]

Here is the question: The price p and the quanitity x sold of a certain product obey the demand equation x=-20p+500 where 0<or equal to x which is <then or equal to 500.

Part A. Express the revenue R as a function of x:
R(x) = -1/20x^2+25x where 0<or equal to x which is <or equal to 500

Part B. What is the revenue if 20 units are sold?
R(20) = (-1/20)(20^2)+(25)(20) = (-0.05)(400)+(500) = $480.00

Is that correct?

Thank you...Cherie

 

[Deleted member 4993]

Here is the question: The price p and the quanitity x sold of a certain product obey the demand equation x=-20p+500 where 0<or equal to x which is <then or equal to 500.

Part A. Express the revenue R as a function of x:
R(x) = -1/20x^2+25x where 0<or equal to x which is <or equal to 500

Part B. What is the revenue if 20 units are sold?
R(20) = (-1/20)(20^2)+(25)(20) = (-0.05)(400)+(500) = $480.00

Is that correct?

Thank you...Cherie

As far as I can tell -- looking good....

 

[stapel]

R(x) = -1/20x^2+25x where 0<or equal to x which is <or equal to 500

R(20) = (-1/20)(20^2)+(25)(20) = (-0.05)(400)+(500) = $480.00

Oh.... Did you mean R(x) to be -(1/20)x[sup:38iw7i2j]2[/sup:38iw7i2j] + 25x, instead of -1/(20x[sup:38iw7i2j]2[/sup:38iw7i2j] + 25x)...?

Eliz.

 

[lilcherbear]

This is what I was thinking:

-(1/20)x2 + 25x

Thank you for your help. Its nice to have re-assurance

Cherie