Profit Equation (need to maximze profit)

  • Context: Undergrad 
  • Thread starter Thread starter Duke77
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on maximizing profit for an automobile manufacturer by determining the optimal pricing strategy for cars. The manufacturer can sell 80,000 cars at a price of $12,000, with sales increasing by 24 units for every $1 decrease in price. The profit function is derived as profit = 296,000p - 24p² - 45,000,000 - 5,500(296,000 - 24p). To find the price that maximizes profit, participants suggest using derivatives or completing the square to identify the vertex of the quadratic profit function.

PREREQUISITES
  • Understanding of profit maximization in economics
  • Knowledge of quadratic functions and their properties
  • Familiarity with derivatives and their application in optimization
  • Basic grasp of fixed and variable costs in manufacturing
NEXT STEPS
  • Learn how to derive profit functions in economics
  • Study the application of derivatives in finding maxima and minima
  • Explore completing the square for quadratic equations
  • Investigate real-world examples of pricing strategies in manufacturing
USEFUL FOR

Economists, business analysts, automotive industry professionals, and anyone involved in pricing strategy and profit optimization.

Duke77
Messages
7
Reaction score
0
An automobile manufacturer finds that 80,000 cars can be sold if each is priced at $12,000. However, the number sold increases by 24 for every $1 decrease in the price. The manufacturer has fixed costs of $45,000,000; in addition, it costs $5,500 to produce each car. How should the cars be priced to maximize profits?

I can do this by trial and error but I would like to know how to set up and do this problem the right way. Thanks.
 
Physics news on Phys.org
Construate the function of profit(price of a car), what means find out how does the profit depend on the price. Then find the maximum of that function using derivatives.
 
"An automobile manufacturer finds that 80,000 cars can be sold if each is priced at $12,000. However, the number sold increases by 24 for every $1 decrease in the price."

Think of this as y= mx+b where if x= 12000, y= 8000. Also "number sold increases --by 24 for every $1 decrease in the price is the same as saying that the slope of the line is m= -24. y= -24x+ b so 8000= -24(12000)+ b. b= 8000+ 24(12000)= 296000.
That is: number sold= 296000- 24(price) or N= 296000-24p.

Of course, the total money brought in is just the number sold times the price:
Np or 296000p- 24p2.

From that, you have to deduct the 45000000 fixed costs and 5500 for each car: a total cost of 450000000+ 5500N= 450000000-5500(296000-24p).

The profit will be: income - cost or
profit= 296000p- 24p2-450000000-5500(296000-24p).

What value of p will make that a maximum?

(I can think of two ways of doing it: set the derivative equal to 0 or, since this is a quadratic, complete the square to find the vertex of the parabola. One of those should be familiar to you.)
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
84K
Replies
2
Views
7K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K