# Total revenue from

• Feb 28th 2008, 05:11 PM
ArmiAldi
Total revenue from
Assume the total revenue from the sale of X items given by R9x) = In (8x+1),
while the total cos to produce x items is C(x)=x/5. Find the approxiamate number of items that should be manufactured so that profit, R (x) - C(x) is maximum.

Thank You all.
• Feb 28th 2008, 05:26 PM
mr fantastic
Quote:

Originally Posted by ArmiAldi
Assume the total revenue from the sale of X items given by R9x) = In (8x+1),
while the total cos to produce x items is C(x)=x/5. Find the approxiamate number of items that should be manufactured so that profit, R (x) - C(x) is maximum.

Thank You all.

Differentiate the function $P(x) = R(x) - C(x) = \ln (8x + 1) - \frac{x}{5}$ 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.
• Feb 29th 2008, 06:35 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Differentiate the function $P(x) = R(x) - C(x) = \ln (8x + 1) - \frac{x}{5}$ 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.

$\frac{dP}{dx} = \frac{8}{8x+1} - \frac{1}{5}$.

$\frac{dP}{dx} = 0 \Rightarrow 0 = \frac{8}{8x+1} - \frac{1}{5} \Rightarrow \frac{8}{8x+1} = \frac{1}{5} \Rightarrow 40 = 8x + 1 \Rightarrow .....$.

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