Marginal profits

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This question is not homework(I'm not even in school). Also, I know how to get the answers but have a question regarding the reasonableness of an answer. This is taken from a calculus book in the chapter introducing the derivative.
Problem: It costs [tex]C(x)=50+\frac{1}{4}x-\frac{1}{30}x^{2}[/tex] dollars to make x dozen donuts in one day [tex]({0}\leq{x}\leq{40})[/tex] and the donuts sell for 75 cents a dozen. a) What are the Revenue R(x) (gross income) and profit P(x) on x dozen donuts? b) What are the average and marginal profits on 30 dozen donuts?
answers a) [tex]R(x)=.75x dollars ; P(x)=\frac{1}{2}x + \frac{1}{30}x^{2}-50dollars[/tex] b)[tex]Averageprofit=-\frac{1}{6}dollar/dozen; Marginal profit=\frac{5}{2}dollars/dozen.[/tex]
My question deals with the marginal profit in part b. It seems that 75 cents per dozen would have to be an upper limit on profit and the $2.50 would not be possible. If one assumes that there are no costs to producing donuts and all the revenue is going into profit, one could not make more than 75 cents/dozen.
Is it that the cost function is completely unrealistic and just created so one can learn the mechanics of taking derivatives?
Thanks for any comments.
 

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  • #2
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How did you get those values for average profit and marginal profit? For instance marginal profit should be a function of x, since we're just differentiating the profit function with respect to x, which is the quantity in terms of dozens donuts. Average profits should just be P(x)/x, provided that x is not zero.
 
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I got them from the answer key in the back of the book. However, when I do the problem over, I end up with a question regarding the answer for average profit. Perhaps this is your question as well.
Marginal profit:
[tex]P'(x)=\frac{1}{15}x+\frac{1}{2}[/tex]

[tex]P'(30)=\frac{5}{2}[/tex]

Average profit:
[tex]\frac{P(x)-P(x_{0})}{x-x_{0}}[/tex]

[tex]\frac{P(30)-P(0)}{30-0}[/tex][tex]=\frac{3}{2}[/tex]

which is not the answer from the book.
It seems that to get its answer one must use 0 instead of P(0)
On one level that might seem reasonable, if one sells 0 donuts, one gets 0 profit, but on another level there is a cost at 0 donuts and the profit may be negative which of course would affect the average.
 
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Redbelly98
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Moderator's note:
Thread moved to Homework area. Homework assignments or any textbook style exercises are to be posted in the appropriate forum in our Homework & Coursework Questions area. This should be done whether the problem is part of one's assigned coursework or just independent study.
 
  • #5
Where did you get your definition of average profit?

Average profit = Total Profit/Quantity
 
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My book defines the average rate of change of a function f(x) as

[tex]\frac{f(x)-f(x_{0})}{x-x_{0}}[/tex]

[tex]x\neq{x_{0}}[/tex]
I used this and replaced P(x) for f(x).
 
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Redbelly98
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My book defines the average rate of change of a function f(x) as

[tex]\frac{f(x)-f(x_{0})}{x-x_{0}}[/tex]

[tex]x\neq{x_{0}}[/tex]
I used this and replaced P(x) for f(x).
Does the question ask for the average profit, or average rate of change of profit?

The average of a function is defined in terms of an integral; average rate of change is as you posted above.
 
  • #8
My book defines the average rate of change of a function f(x) as

[tex]\frac{f(x)-f(x_{0})}{x-x_{0}}[/tex]

[tex]x\neq{x_{0}}[/tex]
I used this and replaced P(x) for f(x).

So that definition would give the 'average rate of change of P(x)' i.e.

change in profit/change in quantity

For average profit which is the same as any average (arithmetic mean) i.e.

total/quantity
 
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Redbelly98: The question asks for "average...profit...". My original post quotes the problem verbatim.
Ellotician: I see your point. Perhaps I was looking for average rate of change instead of average.
Thanks to both of you for your responses.
 
  • #10
Redbelly98
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Would it make sense that "average profit" is the net profit per dozen, i.e. just P(x)/x as snipez said earlier?
answers a) [tex]R(x)=.75x dollars ; P(x)=\frac{1}{2}x + \frac{1}{30}x^{2}-50dollars[/tex] b)[tex]Averageprofit=-\frac{1}{6}dollar/dozen; Marginal profit=\frac{5}{2}dollars/dozen.[/tex]
My question deals with the marginal profit in part b. It seems that 75 cents per dozen would have to be an upper limit on profit and the $2.50 would not be possible. If one assumes that there are no costs to producing donuts and all the revenue is going into profit, one could not make more than 75 cents/dozen.
Is it that the cost function is completely unrealistic and just created so one can learn the mechanics of taking derivatives?
Thanks for any comments.
I am understanding this question better now. The $2.50 is the additional total "profit" you would make by selling one more unit (1 unit = 1 dozen here). If you are losing money, it just means you would lose $2.50 less.
The fact that $2.50 is greater than the $0.75 sale price means that the total cost actually drops when you go from 29 to 30, or from 30 to 31, units. I'm not sure how reasonable that is, but I guess it could happen in theory. In this example, the total cost starts dropping once you get above 4 units.
 

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