Answer:
Step-by-step explanation:
Given that:
R(x) = [tex]\frac{17}{2x+1}[/tex] + 34x − 17
As we know that derivative of revenue function is marginal revenue function .
We will use following rules of derivative
=> dR/ dx = [tex]\frac{-17*2}{(2x+1)^{2} } + 34[/tex]
=> R' (x) = Â [tex]\frac{-34}{(2x+1)^{2} } + 34[/tex]
=> R '(2000) = Â [tex]\frac{-34}{(2*2000+1)^{2} } + 34[/tex] = 34
The revenue when 2000 units are sold is:
R(2000) = [tex]\frac{17}{2*2000+1}[/tex] + 34*2000 − 17 = $69,783
The marginal revenue be "34" and revenue change be "$69,783".
The increase throughout income caused by the acquisition of one more unit of production, is a Marginal revenue. Although marginal income can stay unchanged at a predetermined interval, it's indeed subject to the regulations of decreasing returns as well as will ultimately slow as outcome level grows.
According to the question,
Revenue function,
R(x) = [tex]\frac{17}{2x+1}[/tex] + 34x - 17
Units sold = 2000
(a) The marginal revenue be:
→ [tex]\frac{dR}{dx}[/tex] = [tex]-\frac{17\times 2}{(2x+1)^2}[/tex] + 34
R'(x) = [tex]- \frac{34}{(2x+1)^2}[/tex] + 34
By substituting "x = 2000",
R'(2000) = [tex]- \frac{34}{(2\times 2000+1)^2}[/tex] + 34
       = 34
(b) The revenue change be:
→ R(2000) = [tex]\frac{17}{2\times 2000+1}[/tex] + 34 × 2000 - 17
         = 69,783 ($)
Thus the above response is appropriate.
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