Business Caclulus

[Stylerl01]

I am taking a business calculus class in college and currently seeking assistance that keeps stumbling me.

The question is:
Estimate the price that yields a maximum profit, that is, revenue minus cost.

What's given is:
With x denoting the number of units, a commodity has a demand function modeled by:
p = 50e^(-x/200), x > 0

And a total cost function is:
C = 7x+20

I know how to find revenue and set up the equation, but every time I derive to a solution, my professor gives me the red X.

Thanks again for the assistance.

 

[Deleted member 4993]

I am taking a business calculus class in college and currently seeking assistance that keeps stumbling me.

The question is:
Estimate the price that yields a maximum profit, that is, revenue minus cost.

What's given is:
With x denoting the number of units, a commodity has a demand function modeled by:
p = 50e^(-x/200), x > 0

And a total cost function is:
C = 7x+20

I know how to find revenue and set up the equation, but every time I derive to a solution, my professor gives me the red X.

Thanks again for the assistance.

You have defined 'x' (= number of units). However what are 'p' and C?

How are those related to profit function?

 

[Stylerl01]

Sorry, forgot to add the that the demand function was modeled by:
p = 50e^(-x/200), x > 0

as for C, it is the total cost function which is:
C = 7x + 20

 

[Stylerl01]

Sorry for the double post.

It's related to the profit function in that I need to take revenue (x*p) and minus cost which is the total cost function, C = 7x + 20

 

[JeffM]

Sorry for the double post.

It's related to the profit function in that I need to take revenue (x*p) and minus cost which is the total cost function, C = 7x + 20

I am sorry. I do not know where you are stuck because you have not told us what you have done.

Let n = net income = profit.

Do you think n can be expressed as a function of x and no other variable?
If so, what do you think that function is?
What do you think the derivative of that function is?
At what value or values of x do you think that derivative = 0?

You have not told us any of these things so it is hard to know just where you have gone wrong.

 

[Stylerl01]

Sorry for the double post.

It's related to the profit function in that I need to take revenue (x*p) and minus cost which is the total cost function, C = 7x + 20

I am sorry. I do not know where you are stuck because you have not told us what you have done.

Let n = net income = profit.

Do you think n can be expressed as a function of x and no other variable?
If so, what do you think that function is?
What do you think the derivative of that function is?
At what value or values of x do you think that derivative = 0?

You have not told us any of these things so it is hard to know just where you have gone wrong.

I do apologize, this is the first time I've used a forum like this to seek assistance.

For the maximum profit, which is, revenue minus cost I used the following:

50xe^(-x/200) - 7x + 20
50e^(-x/200) * (1 - x/200) - 7
x = 200

50e^(-200/200)
50e^(-1) = $18.39

And that is what I end up with. So any ideas where I went wrong would be greatly appreciated and thanks again for the patience on my part.

 

[JeffM]

Sorry for the double post.

It's related to the profit function in that I need to take revenue (x*p) and minus cost which is the total cost function, C = 7x + 20

I am sorry. I do not know where you are stuck because you have not told us what you have done.

Let n = net income = profit.

Do you think n can be expressed as a function of x and no other variable?
If so, what do you think that function is?
What do you think the derivative of that function is?
At what value or values of x do you think that derivative = 0?

You have not told us any of these things so it is hard to know just where you have gone wrong.

I do apologize, this is the first time I've used a forum like this to seek assistance.
Hey that's OK. We just need information to figure out where you went wrong?

For the maximum profit, which is, revenue minus cost I used the following:
(Your equation is for profit, not for maximum profit. You will use the derivative of that equation to find where it is maximum.)

50xe^(-x/200) - 7x + 20
No, n = [50xe^(-x/200)] - (7x + 20] = 50xe^(-x/200) - 7x - 20. Minus 20, not plus 20. By the way, it will help you if you use full equations

50e^(-x/200) * (1 - x/200) - 7
I think you are trying to calculate dn/dx here, whuch requires, amomg other things, using the chain rule and the product rule. It might help to take it in steps.
Let u = - x / 200. Let w = e^u. So, n = 50xw - 7x - 20. So what is dn/dx before evaluating dw/dx?

x = 200
I have no idea where this came from. Once you have the derivative, what do you do? Please set it up as an equation so your logic can be folllowed.

50e^(-200/200)
50e^(-1) = $18.39

And that is what I end up with. So any ideas where I went wrong would be greatly appreciated and thanks again for the patience on my part.