Revenue & Profit Functions

[Loganblahtimes2]

"The cost of producing x units of a product is C(x), where C(x) = 20x +100. The product sells for $40 per unit. (a) Find the break even point. (b) What revenue will the company receive if it sells just the number of units from part (a)? (c) Find the profit function."

I'm having trouble on parts B (as I believe I'm overthinking it) and C. Any help is appreciated.

Edit for clarity:
On part B, I believe the equation is 20(5)+100=40(5), which comes out to 200. I feel very unsure of this however and I am looking for a verification and correction if i'm wrong.
On part C I'm unsure of what to do here since the revenue function is needed for the profit function, and the revenue function is not given. I believe the equation for the profit function is 200x-20x-100, which would simplify to 180x-100. Again, I'm looking for a verification and correction if i'm wrong.

 

[Deleted member 4993]

"The cost of producing x units of a product is C(x), where C(x) = 20x +100. The product sells for $40 per unit. (a) Find the break even point. (b) What revenue will the company receive if it sells just the number of units from part (a)? (c) Find the profit function."

I'm having trouble on parts B (as I believe I'm overthinking it) and C. Any help is appreciated.

Edit for clarity:
On part B, I believe the equation is 20(5)+100=40(5), which comes out to 200. I feel very unsure of this however and I am looking for a verification and correction if i'm wrong.
On part C I'm unsure of what to do here since the revenue function is needed for the profit function, and the revenue function is not given. I believe the equation for the profit function is 200x-20x-100, which would simplify to 180x-100. Again, I'm looking for a verification and correction if i'm wrong.

What was the answer for part (a)?

For part (B) you are saying :"....which comes out to 200"

What comes out to 200?

For part (C), what is the definition of profit?

 

[Loganblahtimes2]

What was the answer for part (a)?

For part (B) you are saying :"....which comes out to 200"

What comes out to 200?

For part (C), what is the definition of profit?

Thanks for taking the time to help me!
We are not told anywhere what the revenue function is directly, but I believe it would be 40 as the necklaces sell for $40. I'm confused on if 40 is the revenue function.

If the revenue function is 40 the answer for (A) would be 5.

20(5)+100=40(5) comes out to 200. More specifically, 200=200.

For part C, the definition of profit is not given in the problem. What is put in quotes is the entire problem. According to earlier parts of the course this problem is from, the profit function is "P(x)=R(x)-C(x)".

 

[Steven G]

You seem not to understand what the revenue function gives you. R(x) is the revenue function which tells you how much revenue you take in if you sell x units of goods.

Now if the revenue function was 40 as you said, then R(x) = 40. This means that no matter what x is (the number of units which you sold) the revenue is 40. Do you think that is correct? If not, then what should R(x) equal?

BTW, I do not agree that if the revenue is 40 then the answer to part A is 5.

 

[JeffM]

You did not say what course you are taking. In an economics course, you would learn that

[MATH]\text {revenue = price per unit times units sold, or } r = p * u; [/MATH]
[MATH]\text {expense = fixed expenses plus variable expenses per unit sold times units sold, or } e = f + v * u;[/MATH]
[MATH]\text {and profit = revenue minus expenses, or } \pi = r - e.[/MATH]
And the units sold equal some function d(p).

Now manipulate that algebraically.

 

[Loganblahtimes2]

You did not say what course you are taking. In an economics course, you would learn that

[MATH]\text {revenue = price per unit times units sold, or } r = p * u; [/MATH]
[MATH]\text {expense = fixed expenses plus variable expenses per unit sold times units sold, or } e = f + v * u;[/MATH]
[MATH]\text {and profit = revenue minus expenses, or } \pi = r - e.[/MATH]
And the units sold equal some function d(p).

Now manipulate that algebraically.

I'm still confused about the revenue function. I understand that revenue is equal to (price * units), but how would you convert that into an (R(x)= function?
Thank you for your patience.

 

[HallsofIvy]

I'm still confused about the revenue function. I understand that revenue is equal to (price * units), but how would you convert that into an (R(x)= function?
Thank you for your patience.

In "R(x)" what does the "x" stand for?

You were told that this sells for $40 per unit. If you sell only one unit, your "revenue" ia R(1)= $40. If you sell two units you make R(2)= $40+ $40= 2($40)= $80, If you sell 3 units you make R(3)= $40+ $40+ $40= 3($40)= $120.

If you sell 10 units, how much do you make? What if you sell "x" units? So what is R(x)?

 

[JeffM]

I'm still confused about the revenue function. I understand that revenue is equal to (price * units), but how would you convert that into an (R(x)= function?
Thank you for your patience.

I still do not know whether you have been given this as an example problem in a course in mathematics or in economics. It makes a difference in what kind of detail you need. Context matters. From a pure mathematics perspective, functions can be discussed as abstract things without outside meaning. But if you are studying economics, the functions relate to economic concepts. I am going to assume that you are studying economics because you have not supplied any context.

We have a demand function, normally specified as units sold as a function of price per unit sold. So normally revenue =

[MATH]p * u, \text { where } u = \alpha (p).[/MATH]
So, NORMALLY, the revenue function is a composition of functions of price per unit sold, namely the product of the identity function and the demand function on price. This collapses as a mathematically coherent description if, as is assumed in this problem, the number of units sold is not dependent on price. in such cases, units sold is an independent variable, and the revenue function is a simple function of two variables, namely the product of p and u. The FORMULA for revenue, however, is still the same. (Economists tend to be very sloppy in their use of mathematical vocabulary and have their own weird notation.)

So the revenue function in this problem is simply pu, and the profit function is

[MATH]\pi = pu - f - vu = u(p - v) - f.[/MATH]
In that formula, v is variable expense per unit sold, and f is fixed cost. The breakeven point, the units sold where profit is zero, is then simply

[MATH]0 = \pi = u(p - v) - f \implies u = \dfrac{f}{p - v}.[/MATH]

 

[Loganblahtimes2]

In "R(x)" what does the "x" stand for?

You were told that this sells for $40 per unit. If you sell only one unit, your "revenue" ia R(1)= $40. If you sell two units you make R(2)= $40+ $40= 2($40)= $80, If you sell 3 units you make R(3)= $40+ $40+ $40= 3($40)= $120.

If you sell 10 units, how much do you make? What if you sell "x" units? So what is R(x)?

Oh! I understand now, thank you for clarifying. R(x) = 40x.
Sorry for not specifying which class I'm taking. I'm taking a mathematics class, not economics. The class covers a wide range of many topics in short sections. The section this problem was found in was "Applications of Functions in Business and Economics".

I believe I'm solved the problem. Understand R(x)=40x was what was giving me trouble. That wasn't stated directly anywhere in the course, just demonstrated without reason and I was confused as to where it was being pulled from.

A) For this type of problem, we were told to use C(x)=R(x). In that case, A should be x=5.
20x+100=40x
5=x

B) I simply multiplied the revenue function by 5. B would be $200.
R(x)=40x
40(5)=200

C) For this type of problem (finding the profit function), we were told to use P(x)=R(x)-C(x). The formula would come out to be 20x-100.
40x-[20x+100]
40x-20x-100
20x-100

Thank you everyone!