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:
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)'`
Then, set R'(x) equal to zero and solve for x.
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` .
Since x refers to the number of units sold, round it off to the next higher whole number.
Now that the value of x is known, plug-in this to p=4-lnx .
Hence, 21 units must be sold at $1 each to maximize the revenue.
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