a company estimates that the total revenue, R, in dollars, received from the sale of items is R= ln(1+1000Q^2). Calculate and interpret the marginal revenue if q=10

r=ln(1+1000q^2)

dr/dq=1/(1+1000q^2)*2000q

dr/dq is the "marginal revenue" which is the instantaneous rate of change at q=10, used to approximate the revenue generated by increasing the sales by 1.

The approximation is just that, to truly calculate the additional revenue produced by the next unit sold you would need to find r(11)-r(10)

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MrRoyal MrRoyal · Answer 2 · 2021-11-16T17:00:42+01:00

The marginal revenue at q = 10 is 0.02, and it means that: an increase in quantity by 10 units will result in an output of 0.02

The revenue function is given as:

[tex]\mathbf{R = ln(1 + 1000q^2)}[/tex]

Differentiate, to calculate the marginal revenue function

[tex]\mathbf{R' = \frac{1}{(1+1000q^2)} \times2000q}[/tex]

When q = 10, we have:

[tex]\mathbf{R' = \frac{1}{(1+1000\times 10^2)} \times2000 \times 10}[/tex]

Evaluate the denominator

[tex]\mathbf{R' = \frac{1}{(1+100000)} \times2000 \times 10}[/tex]

Simplify

[tex]\mathbf{R' = \frac{1}{1000001} \times2000 \times 10}[/tex]

Simplify the denominator

[tex]\mathbf{R' = \frac{20000}{1000001} }[/tex]

[tex]\mathbf{R' = 0.02}[/tex]

Hence, the marginal revenue at q = 10 is 0.02, and it means that: an increase in quantity by 10 units will result in an output of 0.02

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