+0  
 
0
474
2
avatar

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
avatar+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
Sort: 

2+0 Answers

 #1
avatar+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
avatar+78750 
0

...........

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

8 Online Users

avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details