total revenue from the sale

20. Assume the total revenue from the sale of x items is given by
R(x) = 39ln (8x+1), while the total cost to produce x items is C(x) = x/5. Findthe approximate number of items that should be manufactured so that profit, R(x) - C(x) , is maximum.

Quote:

Originally Posted by ArmiAldi

20. Assume the total revenue from the sale of x items is given by
R(x) = 39ln (8x+1), while the total cost to produce x items is C(x) = x/5. Findthe approximate number of items that should be manufactured so that profit, R(x) - C(x) , is maximum.

The following is a slightly altered version of the replies given at this thread:

Quote:

Originally Posted by mr fantastic

Differentiate the function $P(x) = R(x) - C(x) = {\color{red}39}\ln (8x + 1) - \frac{x}{5}$ with respect to x. (You know how to differentiate this, right?)

Put that derivative equal to zero to find the x-coordinate of the stationary point. Test the nature of this solution to prove that this stationary point is a maximum turning point.

Then the value of x found is the answer to the question.

Quote:

Originally Posted by mr fantastic

$\frac{dP}{dx} = {\color{red}39} \left( \frac{8}{8x+1} \right) - \frac{1}{5} = {\color{red}\frac{312}{8x+1} - \frac{1}{5}}$ .

$\frac{dP}{dx} = 0 \Rightarrow 0 = \frac{{\color{red}312}}{8x+1} - \frac{1}{5} \Rightarrow \frac{{\color{red}312}}{8x+1} = \frac{1}{5} \Rightarrow {\color{red}1560} = 8x + 1 \Rightarrow .....$ .

The sign test shows that the value of x corresponds to a maximum turning point.

You did not provide any inforamtion concerning what methods might be appropriate. Are you studying the Derivative and its applications? Can you find the first derivative of R(x) - C(x)? That will send you on your way to a solution.