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Mathematics 16 Online
OpenStudy (anonymous):

The cost of controlling emissions at a firm rises rapidly as the amount of emissions reduced increases. Here is a possible model: C(q) = 4,900 + 96q2 where q is the reduction in emissions (in pounds of pollutant per day) and C is the daily cost to the firm (in dollars) of this reduction. What level of reduction corresponds to the lowest average cost per pound of pollutant? (Round your answer to two decimal places.) pounds of pollutant per day What would be the resulting average cost to the nearest dollar? $ per pound

OpenStudy (xapproachesinfinity):

the lowest would be you minimum the way you know it when you studied parabolas

OpenStudy (xapproachesinfinity):

Do you know how to find the minimum or maximum? well in this case it is a minimum since 96 is positive

OpenStudy (anonymous):

yea i think so! dont i just find the derivative and then set it equal to 0 and then plug back in?

OpenStudy (xapproachesinfinity):

eh you learned about derivatives okay you need to find critical points find the min and plug it in you a pretty good idea here^_^

OpenStudy (xapproachesinfinity):

I thought the question was given to you in precalc

OpenStudy (anonymous):

no its business calc

OpenStudy (xapproachesinfinity):

it is the same thing! just you lecture focuses on business models

OpenStudy (anonymous):

so i find the derivative then = 0 right?

OpenStudy (xapproachesinfinity):

yep! now go for it

OpenStudy (anonymous):

i got 0

OpenStudy (anonymous):

192q for the deriv and then =0 and i got 0??

OpenStudy (xapproachesinfinity):

yes you should get f'(0)=0

OpenStudy (xapproachesinfinity):

c'(q) i mean

OpenStudy (anonymous):

so thats my answer?

OpenStudy (xapproachesinfinity):

no!

OpenStudy (anonymous):

or what do i have to do next?

OpenStudy (xapproachesinfinity):

that's just a critical point! you need to find if it is min or not

OpenStudy (xapproachesinfinity):

you need to check if the derivative changes the sign left of 0 and right of 0 if not that wouldn't be a min

OpenStudy (xapproachesinfinity):

well obviously left of zero the derivative is always negative and right of zero it is positive so f(0) is a minimum

OpenStudy (anonymous):

its a min

OpenStudy (anonymous):

right!

OpenStudy (xapproachesinfinity):

f(0) not 0

OpenStudy (xapproachesinfinity):

yes we have a min

OpenStudy (anonymous):

so whats next?

OpenStudy (xapproachesinfinity):

f(0) find the value?

OpenStudy (anonymous):

isnt it just 4900?

OpenStudy (xapproachesinfinity):

yes!

OpenStudy (anonymous):

okay...

OpenStudy (xapproachesinfinity):

the level of reduction is 0 the lowest average that corresponds to that is 4900

OpenStudy (xapproachesinfinity):

you are in good shape to go

OpenStudy (anonymous):

it says its wrong?

OpenStudy (xapproachesinfinity):

eh! let met reread the problem

OpenStudy (anonymous):

mmkay

OpenStudy (anonymous):

anything?

OpenStudy (xapproachesinfinity):

I'm graphing it!

OpenStudy (anonymous):

okay its okay! i actually have to leave to work in a few minutes so ill try and figure it out later:/

OpenStudy (xapproachesinfinity):

eh okay i think you need to solve c(q)=0

OpenStudy (xapproachesinfinity):

good luck

OpenStudy (anonymous):

thank you!

OpenStudy (xapproachesinfinity):

welcome

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