HELP !!!

[tedog1985]

Dont understand how to do this problem at all ..

Assume the total revenue received from the sale of x items is given by

R (x) = 301n(2x + 1),

While the total cost to produce x items is

C(x) = x/2.

Find the number of item that should be manufactured so that teh profit,

P (x) = R(x) - C(x),

Is a maxium.


Can someone help me with this and show me how to do this im lost... Thanks for your time.

 

[Daniel_Feldman]

You need to find when P(x) is max.

Find P'(x).

Set it equal for 0 and solve for x.

Do a test for max/min using factor-line-analysis.

 

[tedog1985]

How do I do this im still lost on how to find p(x) at max how do i find p1(x)

 

[tedog1985]

How do i factor down the R(x) = 301n(2x+1)

 

[stapel]

How do I do this im still lost on how to find p(x) at max how do i find p1(x)

You have provided only P(x), R(x), and C(x). You have not as yet furnished any information on "p(x)" or "p1(x)".

How do i factor down the R(x) = 301n(2x+1)

This already is factored.

Eliz.

 

[Razor X]

Going out on a limb here, are you having trouble finding the derivative of P or R?

P'(x) = R'(x) - C'(x)

R'(x) = 30(2)[1/(2x+1)] <<<chain rule
R'(x) = 60/(2x+1)

 

[tedog1985]

Will someone post a step by step for me.. Im still getting wrong answers and all kinds of stuff thanks so much.

 

[stapel]

Will someone post a step by step for me.

Most of the tutors do not "do" students' work for them. Instead, they try to help students learn how to do the work for themselves.

Im still getting wrong answers

It's good to hear that you've been working on this. Please reply showing what you've tried, and we'll be glad to check it over to help you find any errors.

Thank you.

Eliz.

 

[happy]

Tedog, do you know how to take the derivative of anything? I notice you were having some trouble in your other problems with this concept.

 

[tedog1985]

Ok this is what im getting now. Can someone check me


P(x) = 30ln(2x+1) – x/2
P(x) = ln(2x+1)30 – x/2

dP/dx = 30*2/(2x+1) - ½
dP/dx = 60/2x+1 - ½

60/(2x+1) – ½ = 0
60/(2x+1) = ½

2*20 = (2x+1)/1

40= 2x + 1
40-1 = 2x
39 = 2x
x = 39/2 = 19.5 ~ 20 items since items are not in decimal format

when x = 20items, profit P(20) = ln(2*20+1)30 – 20/2

P(20) = ln(41)30 – 10 = 48.38 – 10 = 38.38.

 

[stapel]

60/(2x+1) = ½

2*20 = (2x+1)/1

I could be missing something, but....

Why did the "1" in the numerator of the right-hand side become the denominator?

Why did the "60" on the left-hand side become a "20"?

Thank you.

Eliz.