Profit maximization help

  • Thread starter zeezey
  • Start date
  • #1
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Homework Statement



daily Cost function C(x) = 5x + 360 -0.001x^2, where x is the number of decks company produces each day and daily cost is in dollars. Suppose that the price that each deck is sold for varies based on the equation given by p(x) = 11.30 - 0.01x, where p is the price per deck in dollars. Find maximum daily profits, price that the company should charge per deck and the maximum daily profit.

The Attempt at a Solution



profit = C(x) * x - P(x)
=x(5x + 360 -0.001x^2) - (11.30 - .01x)
=5x^2 - 360x - 0.001x^3 - 11.30 +0.01x

Can anyone tell me if I set this up correctly? I know I should take the derivative of this but I end up with 2 different answers.
 

Answers and Replies

  • #2
hotvette
Homework Helper
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I believe the 360x term should be +.

Also, one of the two answers will either not make any sense (e.g. negative) or results in a min instead of max. Remember, a zero derivative means min, max, or inflection point. An easy way to check is to plot profit as a function of x and see what the curve looks like.
 
  • #3
11
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Ok, so the two answers I got were 36.96 and 3296.94 ? It seems like 36.96 would works in the equations I need to plug into since I get $10.96 max price and $365.5 max profit but 36.96 total units seems off?
 
  • #4
gb7nash
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profit = C(x) * x - P(x)
This is not right. Ask yourself if this equation makes sense. If I sell apples at 5 dollars a piece and it costs me 3 dollars for each apple, and assuming that we sell 2 apples, plugging into your equation, what do we get? Once you think about this, ignore the equations and ask yourself what is profit? How do we get the amount of money that we make?
 
  • #5
11
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Well I was using the equation Profit = Price * Quantity - Cost . I was just trying to fit this equation into what I was given. I'm not sure how to then if that equation I came up with was incorrect.
 
  • #6
gb7nash
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Profit = Price * Quantity - Cost
This is correct. Unless I'm misreading your other post, what you have written earlier is Cost*Quantity - Price:

profit = C(x) * x - P(x)
 
  • #7
11
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well C(x) is cost yes but in this particular problem it seems to actually be the price equation and P(x) seems to be the cost. in the problem it states that "the price that each deck is sold for varies based on the equation given by P(x) = 11.30 - 0.01x"
 
  • #8
gb7nash
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well C(x) is cost yes but in this particular problem it seems to actually be the price equation and P(x) seems to be the cost.
I'm not really sure what you mean here. "the price that each deck is sold for varies based on the equation" just means that price is given by P(x) = 11.30 - 0.01x. Similarly for cost. What you want to have is P(x)*x - C(x).
 
  • #9
11
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I tried it that way and the total profit per day turned out to be negative?
11.30x - 0.01x^2 - 5x - 360 +0.001x^2
=-0.009x^2 - 6.30x - 360
f'(x) = 0.018x - 6.30 = 0
x = 6.30/.018 = 350 units
-.009(350)^2 - 6.30(350) - 360 = -3667.5.
P(x) = 11.30 - 0.01x = 11.30 - 0.01(350) = $7.80
 
  • #10
gb7nash
Homework Helper
805
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I tried it that way and the total profit per day turned out to be negative?
11.30x - 0.01x^2 - 5x - 360 +0.001x^2
=-0.009x^2 - 6.30x - 360
f'(x) = 0.018x - 6.30 = 0
x = 6.30/.018 = 350 units
-.009(350)^2 - 6.30(350) - 360 = -3667.5.
P(x) = 11.30 - 0.01x = 11.30 - 0.01(350) = $7.80
Your first line is fine:

Profit = 11.30x - 0.01x2 - 5x - 360 +0.001x2

= -.009x2 + 6.30x - 360

not -0.009x2 - 6.30x - 360

Just be careful with your signs. You'll still get 350 for your quantity, but you should now get a reasonable amount when you plug back into the profit equation.
 
  • #11
11
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Oh dang, how did I miss that. Thanks for the help! :)
 

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