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.

Answer 1

Answer: The change in revenue for the sale of 1 more doghouse $ 66.67 dollars

Explanation: Differential is a function that can be used to approximate function value with a great degree of accuracy. This is done by the following.

Mathematical definition of derivative: f'(x) = lim f(x+Δx) - f(x)/Δx.

If Δx is very small:

f'(x) . Δx ≅ f(x+Δx) - f(x)

Knowing that Δy ≅ f(x+Δx) - f(x) and the diferential of variable x can be written by dx as the variable y can be dy:

dy = f'(x) dx

which means that the differential dy is approximately equal to the change Δy, if Δx is very small.

For the question, R(x) = y(x) = 14,000ln(0.01x+1)

f'(x) = $\frac{d[14,000.ln(0.01x+1)]}{dx}$

Using the chain rule, the derivative will be:

f'(x) = 14,000.$\frac{0.01}{0.01x+1}$

dy = 14,000.$\frac{0.01}{0.01x+1}$.dx

dx is the change in x. For the question, the change is 1 (1 more doghouse) and x is 110:

dy = 14,000$\frac{0.01}{0.01.110+1}.1$

dy = $\frac{140}{2.1}$

dy = 66.67

The change in revenue is $66.67 dollars.