Question

A company estimates that the revenue (in dollars) from the sale of x doghouses is given by R(x)=14,000ln(0.01x+1). Use the differential to approximate the change in revenue fro the sale of one more doghouse if 110 doghouses have already been sold. (Round to nearest cent as needed)

Answer 1

Answer:

The change in revenue from the sale of one more doghouse if 110 doghouses have already been sold is dR/dx=66.67 $/doghouse.

Step-by-step explanation:

We have a revenue function that is:

$R(x)=14,000\cdot \text{ln}(0.01x+1)$

We have to approximate the change in revenue from the sale of one more doghouse, if 110 doghouses have already been sold.

That is the marginal revenue at x=110.

The marginal revenue is expressed as the first derivative of the revenue.

Then, we calculate the derivative of R:

$\dfrac{dR}{dx}=\dfrac{d}{dx}[14,000\cdot \text{ln}(0.01x+1)]\\\\\\\dfrac{dR}{dx}=14,000\dfrac{d}{dx}[\text{ln}(0.01x+1)]\\\\\\\dfrac{dR}{dx}=14,000\cdot\dfrac{1}{0.01x+1}\cdot \dfrac{d}{dx}(0.01x+1)\\\\\\\dfrac{dR}{dx}=14,000\cdot\dfrac{1}{0.01x+1}\cdot 0.01\\\\\\\dfrac{dR}{dx}=\dfrac{14,000}{x+100}$

We then evaluate this marginal revenue at point x=110:

$\dfrac{dR}{dx}_{|x=110}=\dfrac{14,000}{110+100}=\dfrac{14,000}{210}=66.67$