The total cost of producing x books is C(x)=50,000+2x dollars, and the total revenue generated by selling x books for R(x)=10x-0.0001x^2 dollars.

The total cost of producing x books is C(x)=50,000+2x dollars, and the total revenue generated by selling x books for R(x)=10x-0.0001x^2 dollars.

1) Find R'(x)

R(x)=10x-0.0001x^2

R'(x)=10-0.0002x

2)Find C'(x)

C(x)=50,000+2x

C'(x)=2

This means that each book costs $2

3)Find d/dx (r(x)-C(x))

d/dx (r(x)-C(x))=10-0.0002x-2

d/dx (r(x)-C(x))=8 - 0.0002x

4) When is the derivative positive, negative?

\(8-0.0002x>0\\ 8>0.0002x\\ 8\div 0.0002>x\\ x<8\div 0.0002\\ x<40000\)

5) To maximize total profits, how many books should be produced? What is the maximum total profit? Explain reason.

I won't keep going as I see CPhill has already answered it.

We already know (1)  and (2)

3)   R(x)  - C( x)   =

10x - 0.0001x^2   - [ 50000 + 2x ]   =

-0.0001x^2 + 8x - 50000          this  the "profit" function

The derivative of this is

-0.0002x + 8

4)   Set the derivative to  0  and solve

-0.0002x  + 8  = 0

8  = 0.0002x      divide both sides by   0.0002

x = 40000    

.Here's the profit function graph : https://www.desmos.com/calculator/ejbztbmci4

The derivative will be positive from about  ( 6834, 40000) and negative from about (40000, 73,166)

5)   40000 books maximizes the profit......and, as the graph shows, the max profit  is $110,000

The maximum profit in this type of problem will always occur at the x value which makes the derivative = 0

How would you get parts 4.) and 5.) without a graph?