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haterofmath
 3 years ago
Rafael estimates that it costs $14 to produce each unit of a
particular commodity that sells for $23 per unit. There is also a
ﬁxed cost of $1,200.
1 Express the cost C (x ), the revenue R (x ) and the proﬁt P (x )
haterofmath
 3 years ago
Rafael estimates that it costs $14 to produce each unit of a particular commodity that sells for $23 per unit. There is also a ﬁxed cost of $1,200. 1 Express the cost C (x ), the revenue R (x ) and the proﬁt P (x )

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gw2011
 3 years ago
Best ResponseYou've already chosen the best response.1C(x)=14x+1200 R(x)=23x P(x)=23x(14x+1200)

haterofmath
 3 years ago
Best ResponseYou've already chosen the best response.0so P(X) isn't 9x1200

gw2011
 3 years ago
Best ResponseYou've already chosen the best response.1The profit=revenuecost. The cost has a variable part (14x) and a fixed part (1200).

gw2011
 3 years ago
Best ResponseYou've already chosen the best response.1In other words, P(x)=R(x)C(x)

haterofmath
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok so how would you find the average profit function AP(x)?

haterofmath
 3 years ago
Best ResponseYou've already chosen the best response.0and smallest number of units needed to produce for a profit?

gw2011
 3 years ago
Best ResponseYou've already chosen the best response.1To answer your question if P(x)=9x1200 is correct, the answer is yes it is correct because if you simplify P(x)=23x(14x+1200), then you get P(x)=9x1200. The average profit is calculated by dividing by the numberof units produced. This comes out to: AP(x)=(9x1200)/x To find the samllest number of units needed to get a profit you do the following: 9x1200=0 9x=1200 x=1200/9 x=133.3=134 Therefore, the smallest number of units needed to get a profit is 134. If you substitute this value into P(x)=9x1200, you get: P(x)=9(134)1200 P(x)=12061200 P(x)=6

haterofmath
 3 years ago
Best ResponseYou've already chosen the best response.0thanks. so do you express R(x) when x units of product are sold at p thousand dollars per unit where p=6x+100

gw2011
 3 years ago
Best ResponseYou've already chosen the best response.1If I understand your question, then I get the following: R(x)=px R(x)=(6x+100)x R(x)=6x^2+100x

haterofmath
 3 years ago
Best ResponseYou've already chosen the best response.0just to verify... how would you express revenue for a product that sells for 110/unit. total cost is fixed overhead of 7500 plus production cost of 60/unit would it be: R(x)= 110 P(x)= 50x+7500

gw2011
 3 years ago
Best ResponseYou've already chosen the best response.1R(x)=110x Profit=RevenueCost P(x)=110x(60x+7500) P(x)=110x60x7500 P(x)=50x7500

haterofmath
 3 years ago
Best ResponseYou've already chosen the best response.0ok. so how would you find P(x) with the price of unit is p=4.2 0.01x and the cost of production is C(x)=0.002x^2 + 30

gw2011
 3 years ago
Best ResponseYou've already chosen the best response.1With p=4.20.01x and C(x)=0.002x^2+30 doesn't look right to me because this will work out to be negatives as follows: Profit=RevenueCosts P(x)=(4.20.01x)(0.002x^2+30) P(x)=4.20.01x0.002x^230 P(x)=0.002x^20.01x25.8 Unless there is something missing, the above indicates that it comes out to being a negative with no profit.

haterofmath
 3 years ago
Best ResponseYou've already chosen the best response.0yea. that's what I got, just wasn't sure about it thanks
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