Did I miss something? The problem is below. I don't care about the final answer, just need help fixing any steps.
Price (p) and quantity (q) share the relationship- p=-.006x+1800
Determine Revenue function, R(x) for selling x units. I have R(x)= # of units*price per unit or R(x)=x*p so R(x)=x(-0.006x+1800)=-0.006x^2+1800x Correct?
Derive a profit function P(x) if cost of prduction is known as C(x)=12,100,000+800x+0.04x^2. I have Profit = Revenue- Cost or P(x)=R(x)-C(x). So P(x)=(-0.006x^2+1800x)-(12,100,000+800x+0.04x^2)= -0.006x^2+1800x-12,100,000-800x-0.04x^2 =-0.046x^2+1000x-12,100,000. Correct?
How many units should be made for max profit? What is the max profit?
For # of units, Max of x is needed. I used the derivitive of P(x) for this as P'(x)=-.092x+100. Set it to 0 so that x=1000/.092 units. Double prime>0 so it's a max. Correct?
Max profit- P(1000/.092)=-.046(1000/.092)^2+1000(1000/.092)-12,100,000 =-66,652,217.40
Is that really right? Neg. profit?
2007-04-03
20:33:45
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4 answers
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asked by
Anonymous
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Science & Mathematics
➔ Mathematics