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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 )

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C(x)=14x+1200 R(x)=23x P(x)=23x-(14x+1200)
so P(X) isn't 9x-1200
The profit=revenue-cost. The cost has a variable part (14x) and a fixed part (1200).

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In other words, P(x)=R(x)-C(x)
oh ok so how would you find the average profit function AP(x)?
and smallest number of units needed to produce for a profit?
To answer your question if P(x)=9x-1200 is correct, the answer is yes it is correct because if you simplify P(x)=23x-(14x+1200), then you get P(x)=9x-1200. The average profit is calculated by dividing by the numberof units produced. This comes out to: AP(x)=(9x-1200)/x To find the samllest number of units needed to get a profit you do the following: 9x-1200=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)=9x-1200, you get: P(x)=9(134)-1200 P(x)=1206-1200 P(x)=6
thanks. so do you express R(x) when x units of product are sold at p thousand dollars per unit where p=-6x+100
If I understand your question, then I get the following: R(x)=px R(x)=(-6x+100)x R(x)=-6x^2+100x
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
R(x)=110x Profit=Revenue-Cost P(x)=110x-(60x+7500) P(x)=110x-60x-7500 P(x)=50x-7500
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
With p=4.2-0.01x and C(x)=0.002x^2+30 doesn't look right to me because this will work out to be negatives as follows: Profit=Revenue-Costs P(x)=(4.2-0.01x)-(0.002x^2+30) P(x)=4.2-0.01x-0.002x^2-30 P(x)=-0.002x^2-0.01x-25.8 Unless there is something missing, the above indicates that it comes out to being a negative with no profit.
yea. that's what I got, just wasn't sure about it thanks
You're welcome

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