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Profit Equation (need to maximize profit) 
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#1
Apr2804, 08:45 AM

P: 7

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
Apr2804, 10:06 AM

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P: 2,751

Maximize the following the following equation for x :
(80,000 + 24x) (12,000  x)  5,500x 


#3
Apr2804, 10:16 AM

HW Helper
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P: 2,327

I like the method you use to understand the material!



#4
Apr2804, 11:19 AM

P: 7

Profit Equation (need to maximize profit)
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
Apr2804, 11:30 AM

P: 354

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 


#6
Apr2804, 11:32 AM

Sci Advisor
P: 2,751

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.



#7
Apr2804, 11:46 AM

P: 7

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


#8
Apr2804, 12:07 PM

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



#9
Apr2804, 12:08 PM

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P: 2,751

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
Apr2804, 12:40 PM

P: 7

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
Apr2804, 12:54 PM

P: 7

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
Apr2804, 01:11 PM

Sci Advisor
P: 2,751

(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. 


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