# Marginal Profit

• February 13th 2010, 09:57 AM
rawkstar
Marginal Profit
A certain item sells for $30. If the cost of producing this item is given by C=.05x^3+100, find the marginal profit when x=10. So i know P(rofit)=R-C and marginal Profit =dP/dx but im not sure how you find R and when you plug in 10 for x and$30

• February 13th 2010, 10:39 AM
tom@ballooncalculus
Revenue will be price times number sold, so...

http://www.ballooncalculus.org/asy/plain/rates.png

... where straight lines differentiate downwards with respect to x.

Spoiler:
http://www.ballooncalculus.org/asy/plain/ratesa.png

Obviously plug in for x now, in the derivative expression.

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• February 15th 2010, 11:43 AM
rawkstar
I don't know how you got the 30x-1/2x^3-100
• February 15th 2010, 11:51 AM
tom@ballooncalculus
Quote:

Originally Posted by rawkstar
I don't know how you got the 30x-1/2x^3-100

Some of it's R and some of it's C...
• February 15th 2010, 11:57 AM
vince
Quote:

Originally Posted by tom@ballooncalculus
Some of it's R and some of it's C...

man Tom...nice style with the spoliers and all that. what prime pedagogy.
• February 15th 2010, 11:59 AM
rawkstar
howd u get R?
• February 15th 2010, 12:01 PM
tom@ballooncalculus
Quote:

Originally Posted by tom@ballooncalculus
Revenue will be price times number sold, so...

.....
• February 15th 2010, 12:05 PM
rawkstar
thanks but shouldnt P=30x-.05x^3-100 since C=.05x^3+100?
• February 15th 2010, 12:15 PM
tom@ballooncalculus
Quote:

Originally Posted by rawkstar
thanks but shouldnt P=30x-.05x^3-100 since C=.05x^3+100?

Gosh, yeah - never saw that zero! (Doh)

Will correct the pics
• February 24th 2015, 12:48 PM
mathidiot123
Re: Marginal Profit
I do not understand where the 1/20 came from. How did you build that equation? And when you plug in x (which is 10), I get -120 and my answer sheet says 15.
• February 24th 2015, 01:51 PM
skeeter
Re: Marginal Profit
Quote:

Originally Posted by mathidiot123
I do not understand where the 1/20 came from. How did you build that equation? And when you plug in x (which is 10), I get -120 and my answer sheet says 15.

note that $0.05 = \frac{1}{20}$

$\frac{dP}{dx} = 30 - 0.15x^2$

evaluated at $x = 10$ ...

$\frac{dP}{dx}_{x=10} = 30 - 0.15(10^2) = 30 - 15 = 15$