hw 11 - chain rule section 2.6: problem 12 (1 point)\na total - revenue function is given by\nr(x)=1400sqrt{x…

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)