Question

hw 11 - chain rule section 2.6: problem 12 (1 point)
a total - revenue function is given by
r(x)=1400sqrt{x^{2}-0.1x}
where (r(x)) is the total revenue, in thousands of dollars, from the sale of (x) airplanes. find the rate at which the total revenue is changing when 20 airplanes have been sold.
dollars per airplane
round to two decimal places.

Explanation:

Answer:

First, find the derivative of \(R(x)=1400\sqrt{x^{2}-0.1x}=1400(x^{2}-0.1x)^{\frac{1}{2}}\) using the chain - rule.

The chain - rule states that if \(y = f(g(x))\), then \(y^\prime=f^\prime(g(x))\cdot g^\prime(x)\).

Let \(u = x^{2}-0.1x\), so \(y = 1400u^{\frac{1}{2}}\).

The derivative of \(y\) with respect to \(u\) is \(y^\prime_{u}=1400\times\frac{1}{2}u^{-\frac{1}{2}} = 700u^{-\frac{1}{2}}\), and the derivative of \(u\) with respect to \(x\) is \(u^\prime_{x}=2x - 0.1\).

Then \(R^\prime(x)=700(x^{2}-0.1x)^{-\frac{1}{2}}\cdot(2x - 0.1)=\frac{700(2x - 0.1)}{\sqrt{x^{2}-0.1x}}\)

When \(x = 20\), we substitute \(x = 20\) into \(R^\prime(x)\):

\[

$$\begin{align*} R^\prime(20)&=\frac{700(2\times20 - 0.1)}{\sqrt{20^{2}-0.1\times20}}\\ &=\frac{700(40 - 0.1)}{\sqrt{400 - 2}}\\ &=\frac{700\times39.9}{\sqrt{398}}\\ &=\frac{27930}{\sqrt{398}}\approx\frac{27930}{19.95}\\ &\approx1399.90 \end{align*}$$

\]

Since \(R(x)\) is in thousands of dollars, the rate of change of the total revenue when 20 airplanes have been sold is \(1399.90\times1000 = 1399900.00\) dollars per airplane.

So the answer is \(1399900.00\)