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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 )
 one year ago
 one year 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 )
 one year ago
 one year ago

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gw2011Best ResponseYou've already chosen the best response.1
C(x)=14x+1200 R(x)=23x P(x)=23x(14x+1200)
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
so P(X) isn't 9x1200
 one year ago

gw2011Best ResponseYou've already chosen the best response.1
The profit=revenuecost. The cost has a variable part (14x) and a fixed part (1200).
 one year ago

gw2011Best ResponseYou've already chosen the best response.1
In other words, P(x)=R(x)C(x)
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
oh ok so how would you find the average profit function AP(x)?
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
and smallest number of units needed to produce for a profit?
 one year ago

gw2011Best ResponseYou've already chosen the best response.1
To 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
 one year ago

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

gw2011Best ResponseYou've already chosen the best response.1
If I understand your question, then I get the following: R(x)=px R(x)=(6x+100)x R(x)=6x^2+100x
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
just 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
 one year ago

gw2011Best ResponseYou've already chosen the best response.1
R(x)=110x Profit=RevenueCost P(x)=110x(60x+7500) P(x)=110x60x7500 P(x)=50x7500
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
ok. 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
 one year ago

gw2011Best ResponseYou've already chosen the best response.1
With 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.
 one year ago

haterofmathBest ResponseYou've already chosen the best response.0
yea. that's what I got, just wasn't sure about it thanks
 one year ago
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