Urgently need help: Max. profit question: Materials for each feeder costs $25, and...

[jc0517]

Materials for each feeder costs $25, and they sell an average of 20 feeders each week, for aprice of $40 each. Market research in the society suggests that for each price reduction of $10, anadditional 5 feeders will be sold. Conversely, for each price increase of $10, 5 fewer feeders will besold.

Without using calculus, find the price per feeder, to the nearest dollar, that the society should chargein order to maximize the profit to the society.

Any help with this question will be much appreciated.
Thank you!

 

[JeffM]

Materials for each feeder costs $25, and they sell an average of 20 feeders each week, for aprice of $40 each. Market research in the society suggests that for each price reduction of $10, anadditional 5 feeders will be sold. Conversely, for each price increase of $10, 5 fewer feeders will besold.

Without using calculus, find the price per feeder, to the nearest dollar, that the society should chargein order to maximize the profit to the society.

Any help with this question will be much appreciated.
Thank you!

You should read "READ BEFORE POSTING."

Is this problem from an economics course? If so, there is an extremely easy way to solve it. According to microeconomic theory, what is the condition necessary for maximization of profits?

If this problem comes from a math course, why does it say to avoid calculus? What are you currently studying?

 

[jc0517]

Hi JeffM,

I am currently studying mathematics for economics.

We have learnt calculus in this course before. I would love to share first how much I have worked through in this question, however, I am just really lost as to how to even begin tackling this one..

 

[mmm4444bot]

I know next to nothing about economics, but here's an algebraic thought.

Let x = the number of $10 reductions in price

Let y = the number of $10 increases in price

Then we have two cases to consider.

With price reductions:

Average number of items sold per week = 20 + 5x

Price per item = 40 - 10x

With price increases:

Average number of items sold per week = 20 - 5x

Price per item = 40 + 10x

If we use the rest of the given information to write functions for profit, we could compare outputs from each case.

Would that lead to any conclusions? :cool:

 

[JeffM]

Hi JeffM,

I am currently studying mathematics for economics.

We have learnt calculus in this course before. I would love to share first how much I have worked through in this question, however, I am just really lost as to how to even begin tackling this one..

The straight-forward mathematical way to address this problem is first to identify the demand function, which has a constant slope of 5 / (- 10) = - 0.5. Moreover d = 20 when p = 40. So

\(\displaystyle 20 = m - 0.5(40) \implies m = 40 \implies d = 40 - 0.5p.\)

Let's check. If p = 40, then d = 20. Good. And

\(\displaystyle d + \Delta d = 40 - 0.5(p + \Delta p) \implies \Delta d = (40 - 0.5p - 0.5 \Delta p) - (40 - 0.5p) = -\ 0.5 \Delta p\).

If price goes up by 10, units sold go down by 5. If price goes down by 10, units sold go up by 5.

Follow that?

Now let's figure out the revenue function.

\(\displaystyle r = p * d = 40p - 0.5p^2.\)

Then the cost function

\(\displaystyle c = 25d = 1000 - 12.5p.\)

Finally the profit function is

\(\displaystyle \pi = r - c = 40p - 0.5x^2 - 1000 + 12.5p = -\ 0.5 p^2 + 52.5p - 1000.\)

Using calculus, you find that profit is maximized where

\(\displaystyle \dfrac{d \pi}{dp} = 0 \implies -\ p + 52.5 = 0 \implies p = 52.5\).

Because the profit function is a parabola, you can find the maximum without calculus.

However, these answers assume you can sell fractional feeders, which is absurd. At a price of 52, you will sell 14 feeders. At a price of 54, you will sell 13 feeders. How do you proceed?