# total revenue from the sale

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• Feb 29th 2008, 06:29 PM
ArmiAldi
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.
• Feb 29th 2008, 06:49 PM
mr fantastic
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.

• Feb 29th 2008, 06:51 PM
TKHunny
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.