# Profit Equation (need to maximize 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?

Thanks for all who help.

2. Apr 28, 2004

### uart

Maximize the following the following equation for x :

(80,000 + 24x) (12,000 - x) - 5,500x

3. Apr 28, 2004

### JasonRox

I like the method you use to understand the material!

4. Apr 28, 2004

### Duke77

Maximize Profit Equation but how do you maximize profits

I appreciate the help with the equation and I hope to figure out how to maximize profits by myself but at this point it's a trial and error system. I know there must be an easier way to do this. Thanks all.

5. Apr 28, 2004

### PRodQuanta

graph and find the feasible region on your graph. Pick out your vertex points. One will be the maximum. Take this and find an equation for it.

6. Apr 28, 2004

### uart

What grade are you in Duke. Have you done any calculus (differentiation) yet. Alternatively have you studied the properites of parabola's before ? Either of those things will give you a "way in" to maximizing that equation.

Last edited: Apr 28, 2004
7. Apr 28, 2004

### Duke77

I am a freshman in college and am currently finishing up calculus, my math teacher gave us this problem to work on in groups but since I was gone I am trying to figure out everything by myself and he refuses to help anyone. I have done differentiation but I forget how to do it since it was months ago. Could you help me? If not I understand but I'm trying to complete this by 5 pm tonight.

One other thing, I tweaked your equation so it worked

[(80,000+24x)(12,000-x)-(5,500)(24x)+(80,000)(5,500)]-45,000,000=profit

Last edited: Apr 28, 2004
8. Apr 28, 2004

### NateTG

Hint: The maximum on an interval either occurs at the end of the interval, or at a point where the derivative is zero.

9. Apr 28, 2004

### uart

Just expand (multiply out the bracketed terms) that profit equation and you'll see it a parabola. You should get -ax^2 + bx + c. The x^2 term is negative so it's a upsidedown parabola which has it's maximum were the slope (derivative) is zero. So just differentiate the equation and find the value of x that makes the derivative zero.

10. Apr 28, 2004

### Duke77

I multiplied out the whole equation and got -24x2+76,000x+475,000,000 and when i put it in my calculator i get an error "window range". What should i do?

11. Apr 28, 2004

### Duke77

Disregard my last post, I figured it out and thx so much for everyone who helped me. You saved me lots of stress. Thanks.

I took the derivative and solved for it when it was 0. I got 1583.3 for x to maximize profits if anyone was curious.

12. Apr 28, 2004

### uart

OK that's good. BTW I deliberately omited the "constant" parts of the cost equation becuase they make no difference to the resulting maximization. I did however make a mistake in the original equation as it should have been (omiting constant costs),

(80,000 + 24x) (12,000 - x) - 5,500 * 24x as the "x" dependant terms.

So expanding this out gives, -24x^2 + 76,000x + constant

The constant doesnt matter, the derivative is -48x + 76,000 and all you've got to do is work out what value of "x" makes that zero.