Profit Equation (need to maximize profit)

[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.

 

[uart]

Maximize the following the following equation for x :

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

 

[JasonRox]

I like the method you use to understand the material!

 

[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.

 

[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.

Paden Roder

 

[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.

 

[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

 

[NateTG]

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

 

[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.

 

[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?

 

[Duke77]

Disregard my last post, I figured it out and thanks 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.

 

[uart]

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

OK that's good. BTW I deliberately omited the "constant" parts of the cost equation because 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" dependent terms.

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

The constant doesn't matter, the derivative is -48x + 76,000 and all you've got to do is work out what value of "x" makes that zero.