# Each of the functions is a company's price function, where p is the price (in dollars) at which quantity x (in thousands) will be sold. 1. Find the revenue function R(x). 2. Find the quantity and...

Each of the functions is a company's price function, where p is the price (in dollars) at which quantity x (in thousands) will be sold. 1. Find the revenue function R(x). 2. Find the quantity and price that will maximize revenue. When p=4-ln x

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(1) To determine the revenue function, apply the formula:

Revenue= Price per unit * Number of units sold

Since the price per unit is p and the number of units sold is x, then:

`R(x)=px`

Plug-in p=4-lnx.

`R(x)=(4-lnx)x`

Hence, the revenue function is` R(x)=(4-lnx)x ` .

(2) To solve for the values of x and p that will maximize the revenue,take the derivative of R(x). Apply the product rule which is `(u*v)=u*v'+v*u` .

`R'(x)=(4-lnx)*x' + x*(4-lnx)'`

`R'(x)=(4-lnx)*1+x(-1/x)`

`R'(x)=4-lnx-1`

`R'(x)=3-lnx`

Then, set R'(x) equal to zero and solve for x.

`0=3-lnx`

`0+lnx=3-lnx+lnx`

`lnx=3`

And convert this to exponential equation to get x only at one side of the equation.
Note that the equivalent exponential form of `ln m = a`  is ` m=e^a` .

`x=e^3`

`x=20.09`

Since x refers to the number of units sold, round it off to the next higher whole number.

`x=21`

Now that the value of x is known, plug-in this to p=4-lnx .

`p=4-ln21`

`p=1`

Hence, 21 units must be sold at \$1 each to maximize the revenue.