# marginal analysis and differentials

I don't understand this question hope somebody can help me.

Suppose that the cost to produce an LCD computer monitor is $\$75$. Furthermore, suppose that when the selling price is p dollars, the number of monitors that can be sold at that price is$1000e^{-p/50}$. Find and interpret the marginal profit and the marginal average profit of producing the 61st LCD monitor (You will need to express R p.q in terms of q alone) Annuities: When regular payments of P dollars are made n times per year and earn interest at a rate of r*100% per year compounded n times per year. then after t years of payments the accumulated value of the payments is A=p[[{1+r/n}^(n*t)-1]/(r/n)] Find the expected value ofA based on making n=1 payment(per year)of P$1000 over t=20 years earning interest at a rate of 4% (r=0.04).

use differentials to estimate the range of possible values of A if r can vary by as much as a quarter point (0.25%)

Revenue = Price * Quantity. but quantity depends upon price.

$Q = 1000 e^{-\frac P{50}}\\ \frac {dQ}{dP} = -\frac {Q}{50}$

$MR = \frac {dR}{dQ} = {dP}{dQ} Q + P\\ \frac {dP}{dQ} = \frac{1}{\frac{dQ}{dP}}\\ MR = -50+P$

How about in terms of Q? $P = 50(\ln 1000 - \ln Q)\\ MR = 50(\ln 1000 - \ln Q - 1)$

The $61^{st}$ monitor? $R(60) - R(61) = MR(60.5) = \$90.2$Marginal profit = Marginal revune - marginal costs =$15.20, the firm would be more profitable it it increased output.

$\texttt{average profit = (Revenue - costs)}/Q = (QP - C)/Q$

Marginal Average Profit = $\frac {dP}{dQ} + \frac {C}{Q^2}\\ \frac {-50Q + 75}{Q^2}$

And, I have never had a use for "marginal average profit."