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pshichka [43]
3 years ago
7

There are two routes for driving from A to B. One is freeway, and the other consists of local roads. The benefit of using the fr

eeway is constant and equal to 3.6, irrespective of the number of people using it. Local roads get congested when too many people use this alternative, but if not enough peoople use it, the few isolated drivers run the risk of becoming victims of crimes. Suppose that when a fraction of x of the population is using the local roads, the benefit of this mode to each driver is given by:
1 + 8x – 9x²

Required:
a. Draw a graph showing the benefits of the two driving routes as functions of x, regarding x as a continuous variable between 0 and 1.
b. Identify all possible equilibrium traffic patterns from your graph in part (a). Which equilibria are stable? Which ones are unstable? Explain.
c. What value of x maximizes the total benefit to the whole population?
Business
1 answer:
STALIN [3.7K]3 years ago
5 0

Answer:

a) attached below

b) stable equilibria = x = 0.1 , x = 0.8

   unstable equilibria = other value except 0.1 , 0.8

c) 0.5 , 0.6

Explanation:

Benefit of using the local roads = 1 + 8x - 9x^2

Benefit of using the free way = 3.6

a) Attached below is the required graph

<u>b) Determine The possible equilibrium traffic patterns from the graph </u>

stable equilibria : x = 0.1 ,  x = 0.8 ( this id because at these given value the benefits of using either routes is equal )

unstable equilibria :  every other value of X except 0.1 and 0.8

<u>c) Determine the value of x that maximizes the total benefit to the population</u>

The value of X that maximizes the total benefit to the population = 0.5 and 0.6

attached below is the detailed solution

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Answer:

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Explanation:

For this case we have the following function for the sales

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Part a          

For this case we want to find the derivate of S respect to t and we got:

\frac{dS}{dt}=S'(t)= 2(0.5) t +3 = t+3

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S(4) = 0.5*(4^2) +3(4) -5=15 million

Part c

For this case we have the total sales $ 15 millions after t =4 months

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S'(4) = 4+3 = 7 million

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