# Profit Equation (need to maximze profit)

1. Apr 28, 2004

### Duke77

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.

2. Apr 28, 2004

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.

3. Apr 28, 2004

### HallsofIvy

Staff Emeritus
"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.)