Question

A company estimates that the revenue​ (in dollars) from the sale of x doghouses is given by ​R(x) = 10,000 ln (0.01 x + 1). Use the differential to approximate the change in revenue from the sale of one more doghouse if 120 doghouses have already been sold.

Answer 1

Answer:

$45.45

Step-by-step explanation:

Data provided in the question:

Revenue function, R(x) = 10,000 ln(0.01 x + 1)

now,

change in revenue from sales doghouses i.e, $\frac{d\textup{R}}{d\textup{x}}$

$\frac{d\textup{R}}{d\textup{x}}$  = $10,000\times\frac{\textup{1}}{\textup{0.01x + 1}}\times(0.01)$

because $\frac{d\textup{ln(A)}}{d\textup{x}}$=$\frac{\textup{1}}{\textup{A}}\times\frac{dA}{dx}$  

at x = 120

$\frac{d\textup{R}}{d\textup{x}}$  = $10,000\times\frac{\textup{1}}{\textup{0.01(120) + 1}}\times(0.01)$

or

$\frac{d\textup{R}}{d\textup{x}}$  = $10,000\times\frac{\textup{1}}{\textup{1.2 + 1}}\times(0.01)$

or

$\frac{d\textup{R}}{d\textup{x}}$  = $10,000\times\frac{\textup{1}}{\textup{2.2}}\times(0.01)$

or

$\frac{d\textup{R}}{d\textup{x}}$  = $45.45

hence,

the change in revenue from the sale of one more doghouse if 120 doghouses have already been sold is $45.45