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Help with equation please......

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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=

Guest Oct 27, 2015

Best Answer

#1
+1792
+10

$$\mbox{I'm going to assume it's linear and then check.}\\ (p-32.50)=m(N-250)\\ m=\dfrac{31-32.5}{400-250} = \dfrac{-1.5}{150}=-\dfrac 1 {100} \\ \mbox{so }(p-32.50)=-\dfrac 1 {100}(N-250)\\ p=-\dfrac{N}{100}+35$$

$$\mbox{Checking the two inside points.}\\ 32=-3+35 \Rightarrow True\\ 31.50=-3.5+35 \Rightarrow True\\ \mbox{so all the points check out and the formula is correct.}$$

(b) $$R(N)=N p(N) = N\left(35 - \dfrac N {100}\right)$$

(c) $$Profit(N) = R(N) - C(N) = N\left(35-\dfrac N{100}\right)-(30N+700)$$

$$Profit(N)=5N - \dfrac{N^2}{100}-700$$

Rom  Oct 28, 2015
edited by Rom  Oct 28, 2015
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2+0 Answers

#1
+1792
+10
Best Answer

$$\mbox{I'm going to assume it's linear and then check.}\\ (p-32.50)=m(N-250)\\ m=\dfrac{31-32.5}{400-250} = \dfrac{-1.5}{150}=-\dfrac 1 {100} \\ \mbox{so }(p-32.50)=-\dfrac 1 {100}(N-250)\\ p=-\dfrac{N}{100}+35$$

$$\mbox{Checking the two inside points.}\\ 32=-3+35 \Rightarrow True\\ 31.50=-3.5+35 \Rightarrow True\\ \mbox{so all the points check out and the formula is correct.}$$

(b) $$R(N)=N p(N) = N\left(35 - \dfrac N {100}\right)$$

(c) $$Profit(N) = R(N) - C(N) = N\left(35-\dfrac N{100}\right)-(30N+700)$$

$$Profit(N)=5N - \dfrac{N^2}{100}-700$$

Rom  Oct 28, 2015
edited by Rom  Oct 28, 2015
#2
+78750
0

...........

CPhill  Oct 28, 2015
edited by CPhill  Oct 28, 2015

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