Homework Help

Production Cost The price p, in dollars, of a certain product and the quantity x sold...

user profile pic

ms718 | Student | eNoter

Posted April 28, 2013 at 4:46 AM via web

dislike 0 like

Production Cost The price p, in dollars, of a certain product and the quantity x sold obey the demand equation

P=-1/4x+100, 0 ≤ x ≤ 400

Suppose that the cost C, in dollars, of producing x units is c=`sqrt(x)`/25+600

Assuming that all items produced are sold, find the cost C as a function of the price p.

3 Answers | Add Yours

user profile pic

mvcdc | Student, Undergraduate | (Level 1) Associate Educator

Posted April 28, 2013 at 5:07 AM (Answer #1)

dislike 2 like

We are given the following:

`P = -1/4x + 100`

`C=sqrt(x)/25 + 600`

 

We want to express cost, C, as a function of Price. First, we isolate x from the first equation:``

`P = -1/4 x + 100`

`-4P = x - 400`

`x = 400 - 4P`

Then, we substitute this expression for x to the cost function:

`C = sqrt(400-4P)/4 + 600`

`C=sqrt(4*(100-P))/4 + 600`

`C = 2sqrt(100-P)/4 + 600`

`C = sqrt(100-P)/2 + 600`

Next, note that x in the very first equation (expression for P) only ranges from 0 to 400, including 0 and 400. Therefore, P in the final expression for C ranges from 0 to 100. (P = 0 when x is 400, and P = 100 when x is 0).

user profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 28, 2013 at 5:31 AM (Answer #2)

dislike 2 like

The price of the product p in terms of the quantity sold x is:

`p=(-1/4)*x+100, 0 <= x <= 400` . This can be rewritten as:

`(p - 100) = (-1/4)*x`

=> `x = 400 - 4p`

The cost of production is `c = sqrt x/25 + 600`

Substituting `x = 400 - 4p`

=> `c = sqrt(400 - 4p)/25 + 600`

=> `c = (2/25)*sqrt(100 - p) + 600`

The cost of the product in terms of its price is `c = (2/25)*sqrt(100 - p) + 600`

user profile pic

ms718 | Student | eNoter

Posted April 28, 2013 at 5:56 AM (Answer #3)

dislike 0 like

We are given the following:

 

We want to express cost, C, as a function of Price. First, we isolate x from the first equation:

Then, we substitute this expression for x to the cost function:

Next, note that x in the very first equation (expression for P) only ranges from 0 to 400, including 0 and 400. Therefore, P in the final expression for C ranges from 0 to 100. (P = 0 when x is 400, and P = 100 when x is 0).

 

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes