In general, the highest price p per unit of an item at which a manufacturer can sell N items is not constant but is, rather, a function of N. Suppose the manufacturer of widgets has developed the following table showing the highest price p, in dollars, of a widget at which N widgets can be sold.
(a) Find a formula for p in terms of N modeling the data in the table.
p=
Number N Price p
250 32.50
300 32.00
350 31.50
400 31.00
(b) Use a formula to express the total monthly revenue R, in dollars, of this manufacturer in a month as a function of the number N of widgets produced in a month.
R =
(c) On the basis of the tables in this exercise and using cost, C = 30N + 700, use a formula to express the monthly profit P, in dollars, of this manufacturer as a function of the number of widgets produced in a month.
p=
+6248 \(\mbox{I'm going to assume it's linear and then check.}\\ (p-32.50)=m(N-250)\\ m=\dfrac{31-32.5}{400-250} = \dfrac{-1.5}{150}=-\dfrac 1 {100} \\ \mbox{so }(p-32.50)=-\dfrac 1 {100}(N-250)\\ p=-\dfrac{N}{100}+35\)
\(\mbox{Checking the two inside points.}\\ 32=-3+35 \Rightarrow True\\ 31.50=-3.5+35 \Rightarrow True\\ \mbox{so all the points check out and the formula is correct.}\)
(b) \(R(N)=N p(N) = N\left(35 - \dfrac N {100}\right)\)
(c) \(Profit(N) = R(N) - C(N) = N\left(35-\dfrac N{100}\right)-(30N+700)\)
\(Profit(N)=5N - \dfrac{N^2}{100}-700\)
+6248 \(\mbox{I'm going to assume it's linear and then check.}\\ (p-32.50)=m(N-250)\\ m=\dfrac{31-32.5}{400-250} = \dfrac{-1.5}{150}=-\dfrac 1 {100} \\ \mbox{so }(p-32.50)=-\dfrac 1 {100}(N-250)\\ p=-\dfrac{N}{100}+35\)
\(\mbox{Checking the two inside points.}\\ 32=-3+35 \Rightarrow True\\ 31.50=-3.5+35 \Rightarrow True\\ \mbox{so all the points check out and the formula is correct.}\)
(b) \(R(N)=N p(N) = N\left(35 - \dfrac N {100}\right)\)
(c) \(Profit(N) = R(N) - C(N) = N\left(35-\dfrac N{100}\right)-(30N+700)\)
\(Profit(N)=5N - \dfrac{N^2}{100}-700\)
