Answer to Question #92065 in Algebra for linh

Question

In order to widen a section of the Pacific Highway from three lanes to four lanes in preparation for the holidays, a construction company will have to close one of the existing lanes for a period of one week.

The Department of Transport would like to have the traffic move as quickly as possible along the one available lane, but does not want a lot of rear-end collisions. They need to consider what speed limit should be set in order to maximise the flow of traffic and still ensure safe travel.

The Department of Transport Engineers have produced the following formula for traffic flow,
F=(0.27 ̅V)/(4.2+0.1905V+0.006V^2 )

where F is the number of cars per second passing a given point, and V is the common speed in km/h at which all cars travel to provide uniform traffic flow.

By using differentiation, determine the speed limit that should be posted on the road to maximise the traffic flow. State any assumptions that affect your results.
Assuming that the cars stick to the advised speed.

Answer 1

Using the quotient rule for differentiation (namely (f/g)' = (f'g -f g')/ g²), we obtain:

F'(V) =[(0.27 *V)/(4.2+0.1905V+0.006V²)]'=

=[0.27*(4.2+0.1905V+0.006V²) - 0.27*V*(0.1905 + 0.012V]/ (4.2+0.1905V+0.006V²)² =

= (1.134 + 0.00162* V² - 0.00324V²) / (4.2+0.1905V+0.006V²)² =

= (1.134 - 0.00162V²) / (4.2+0.1905V+0.006V²)²

F'(V) =0 => 1.134 - 0.00162*V² =0 => V² = 1.134 / 0.00162 => V² = 700 =>

$$V=\sqrt{700} = 26.46 km/h$$

(we have ignored the negative root since that is out of scope).

When V is between the two roots $-\sqrt{700},\sqrt{700},$ the 1st derivative is positive, so F is strictly increasing.

When $V\geq \sqrt{700},$ the 1st derivative becomes negative and F is decreasing.

Therefore, the maximum value for F is obtained when

$$V=\sqrt{700},$$

which means the maximum speed is 26.46 km/h