# Price Optimization problem

1. Dec 7, 2012

### domyy

1. The problem statement, all variables and given/known data

A wholesale paint dealer is buying and distributing x cases of paint per week. She incurs the following expenses:

(1) Fixed costs of $1200 (2) An expense of$60x per week representing the cost of x cases to the dealer ($60 per case) (3) A cost of$x^2/24 per week for storing the inventory, handling accounts, etc.

Sales can be maintained at a rate of x cases per week at a price p dollars per case, where x=2160-24p. Due to space and other limitations, the dealer's maximum level of operation is 1000 cases per week.

a) Determine the price p at which the dealer should sell each case to maximize the weekly profit.

2. Relevant equations

Would it be correct:

p(x) =x.p- 60x - 1200 - x^2/24

I was wondering if in (x.p) x would be 1000 because it's the maximum number of cases per week.

Last edited: Dec 7, 2012
2. Dec 7, 2012

### Ray Vickson

Is p(x) supposed to be weekly profit? If so, it is wrong.

As I explained in a previous thread, you need to proceed carefully and systematically. Let me one more time outline the types of steps you should take to avoid errors.

Do you want to use x as your variable? If so, what is the selling price p? (The question asks for p, but you can use x instead and then get p.) In terms of x, what is the weekly revenue from sales? In terms of x, what are the weekly costs? Please answer these questions before taking the next step---seriously!

After doing the above steps, you should put them together to get an expression for the weekly profit. Now you have a well-defined optimization problem, and solving it is a separate issue, but not excessively difficult.

Finally, never use the exact same symbol (p in this case) to stand for two totally different things in the same problem. Choose another symbol for profit, such as P for example. (This is OK because P and p are not *exactly* the same---unless you mis-type one of them.)

3. Dec 7, 2012

### BruceW

not necessarily. How are you going about this problem? It has asked you to maximise profit. So first, write down the equation for profit

Edit: this is in reply to domyy's changed post.

4. Dec 7, 2012

Yes, I want to use x as my variable. The selling price is $60. The weekly revenue would be x.p. The cost are 60x + 1200 + x^2/24 5. Dec 7, 2012 ### Ray Vickson Write out the weekly revenue in detail. 6. Dec 7, 2012 ### domyy Hi, BruceW. I was considering profit = [x.p (revenue)] - [60x + 1200 + x^2/24 (costs)]. Is it wrong? 7. Dec 7, 2012 ### domyy Would weekly revenue = [(1000 - 60x)x] be correct then ? Last edited: Dec 7, 2012 8. Dec 7, 2012 ### Ray Vickson You have an expression for revenue; it involves x (the number sold per week) and the selling price per unit. If you read the question *carefully and completely* you will get the answers you need. I really do believe that part of your difficulty is caused by your NOT reading the question with care, and jumping far too quickly to wrong conclusions. 9. Dec 7, 2012 ### domyy "Sales can be maintained at a rate of x cases per week at a price p dollars per case, where x=2160-24p. Due to space and other limitations, the dealer's maximum level of operation is 1000 cases per week." Here, it says x cases, price p. And the maximum level of operations is 1000. However, it does not mean I have 1000 cases every week. So, maybe I should write (x-60x)x ? Last edited: Dec 7, 2012 10. Dec 7, 2012 ### Ray Vickson You keep "guessing" instead of thinking it through. To calculate revenue you need p and x. You are using x as your variable, so that for any numerical value of x, you calculate cost and revenue---but to do that you need p as well. So, in terms of x, how do you get p? As I said: don't guess; you don't need to. You are told everything you need. 11. Dec 7, 2012 ### domyy Ok. I tried to solve the problem again and this is what I got: Revenue: x*p Being p = -x + 2160/24 Revenue: x*(-x+2160/24) Revenue will then be: -x^2 + 2160x/24 Profit: revenue - cost Cost: (1200 + 60x + x^2/24) Profit will then be: (-x^2 + 2160x/24) - (1200 + 60x + x^2/24) => (-x^2 + 2160x - 28800 + 1440x + x^2)/24 => (720x - 28800)/24 => 30x - 1200 To maximize profit: p'(x) = 30x - 1200 = 30 C = (1200 + 60x + x^2/24 )/x = (1200 + 60(30) + (30)^2/24)/30 = [(28800 + 43200 + 900)/24] : 30 = 72900/24(30) =$101.25

Last edited: Dec 7, 2012
12. Dec 7, 2012

### domyy

No response? Oh please, don't tell me I am wrong again :grumpy:

13. Dec 8, 2012

### Ray Vickson

p is not given by
$$p = -x + \frac{2160}{24}$$ which is what you wrote; it is given by
$$p = \frac{-x + 2160}{24}.$$
You need parentheses, like this: p = (-x + 2160)/24 = -(x/24) + 2160/24.

Also, I don't understand why you set p'(x) equal to 30 (which is what you wrote above) Shouldn't you set p'(x) to zero? Also, the expression expression for profit is not what you wrote, so yes---you made an important error. Go back and do it again. Finally, do not use the same letter p (as in p'(x)); you have already reserved the letter p for something else, as I have explained to you before.

I give up; this is my last message.

14. Dec 8, 2012

### BruceW

keep trying though dude, you're getting closer.

15. Dec 17, 2012

### domyy

The expression for profit will be: (-2x^2 + 720x - 28800)/24 correct?

[(-2x^2 + 720x - 28800)/24]' = 0 ?

Will that be correct?

Last edited: Dec 17, 2012
16. Dec 18, 2012

### BruceW

That is exactly correct. Nice work man :) You're almost there now

17. Dec 18, 2012

### domyy

Thank you so much!!!!! :)