Question

The average number of vehicles waiting in line to enter a parking lot can be modeled by the function f (x )equalsStartFraction x squared Over 2 (1 minus x )EndFraction ​, where x is a number between 0 and 1 known as the traffic intensity. Find the rate of change of the number of vehicles waiting with respect to the traffic intensity for the intensities ​(a) xequals0.3 and ​(b) xequals0.6.

Answer 1

Answer:

The rate of change of the number of vehicles waiting with respect to the traffic intensity for the intensities are:

a) 0.52

b) 2.63

Step-by-step explanation:

The rate of change of a function at a given point P can be obtained by evaluating the 1st derivative of the function in P. Thus,

$f(x)=\frac{x^2}{2(1-x)}$

$\frac{df(x)}{dx} = \frac{d(\frac{x^2}{2(1-x)})}{dx} \\\frac{df(x)}{dx} =\frac{x}{1-x} + \frac{x^2}{2(1-x)^2} \\\frac{df(x)}{dx} = -\frac{x(x-2)}{2(x-1)^2}$

Now for we just need to evaluate in each of the given points

a) $\frac{df(0.3)}{dx} =-\frac{0.3(0.3-2)}{2(0.3-1)^2}\\\boxed{\frac{df(0.3)}{dx} =0.520408 \approx 0.52}$

b)$\frac{df(0.6)}{dx} =-\frac{0.6(0.6-2)}{2(0.6-1)^2}\\\boxed{\frac{df(0.6)}{dx} = 2.625\approx 2.63}$