Applied Maxima and Minima

In summary, the problem involves finding the profit-maximizing price for a monopolist's product, given a demand equation and average-cost function. The solution involves setting marginal cost equal to marginal revenue and solving for the quantity. After substituting the quantity into the demand equation, the profit-maximizing price is found to be 5.
  • #1
jose_m
2
0

Homework Statement


For a monopolist's product, the demand equation is:

p = 42 - 4q

and the average-cost function is

c = 2 + (80/q)

Find the profit-maximizing price.


Homework Equations


When i started to solve the problem, i deduced from what i needed to find that I needed to make up a function, plugging in the above functions into profit = (price-cost)quantity.
To what extent I'm correct I'm not sure.

The Attempt at a Solution


I tried plugging it in like this: profit = ((42-4q-2-(80/q)) all that multiplied by q. I don't seem to know if i multiply that whole equation just by q, or by finding a formula for q from those other functions they already gave me. I tried it and it gave me q= 80/(c-2) but then, do i have to substitute c in that formula.

I really don't know where to go from here. would appreciate any help given please.
 
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  • #2
At maximimum profit, demand= cost. I would have thought that was one of the frst thing you would have learned!
 
  • #3
go it solved! thanks for that little fact i did miss. that equals to 4q^2-40q+80, which its derivative is 8q-40, and its critical value is 5, with a maximum of 5. thanks. :D
 
  • #4
Since this is a monopoly problem, you should be looking to equate marginal cost (MC) to marginal revenue (MR).

Total cost is:

C = Average Cost times q = cq = 2q + 80

MC = C' = 2.

Total revenue is:
R = pq = 42q - 4q^2.

MR = R' = 42 - 8q

MC = MR ===> 2 = 42 - 8q ===> 8q = 40 ===> q = 5, so your answer is correct.
 
Last edited:

1. What is the concept of applied maxima and minima?

Applied maxima and minima are mathematical tools used to determine the highest and lowest values of a function within a given range. They are commonly used in optimization problems to find the most efficient solution.

2. How are maxima and minima different from each other?

Maxima refers to the highest value of a function, while minima refers to the lowest value. In other words, maxima is the peak or crest of a function, while minima is the valley or trough.

3. What are some real-world applications of applied maxima and minima?

Applied maxima and minima are used in various fields such as economics, engineering, and physics. They can be used to optimize production processes, determine the most profitable price for a product, or find the optimal design for a structure.

4. How are maxima and minima calculated?

Maxima and minima can be calculated using calculus. The first derivative of a function is set to zero, and the resulting equation is solved to find the critical points. The second derivative is then used to determine whether these points are maxima or minima.

5. Are there any limitations to using applied maxima and minima?

Yes, there are some limitations to using applied maxima and minima. These methods may not work for all functions, especially those with multiple peaks and valleys. Additionally, the accuracy of the results depends on the precision of the input data and the assumptions made in the calculations.

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