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n200080 [17]
3 years ago
6

Suppose the supply curve of portable radio rentals in Golden Gate Park is given by P = 5 + 0.1Q, where P is the daily rent per u

nit in dollars and Q is the volume of units rented in hundreds per day. The demand curve for portable radios is P = 20 – 0.2Q. a. If each portable radio imposes $3 per day in noise costs on others, by how much will the equilibrium number of portable radios rented exceed the socially optimal number?
Mathematics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

10 or 1000 in hundreds of units

Step-by-step explanation:

Given the supply curve of portable radio rentals:

P = 5 + 0.1Q

and the demand curve for portable radio rentals:

P = 20 - 0.2Q

we need to find the equilibrium of these two curves ( replace either's right hand side on the other's left hand side)

we have:

20 - 0.2Q = 5 + 0.1Q

solving we have Q=50 or 5000 in hundreds of units

Now the noise is equal to 3, therefore, we have to find the social supply curve by adding 3 to the first supply curve. following:

P = 8 + 0.1Q

We find the intersection bewteen the social supply curve and the demand curve ( social equilibrium rental ):

20 - 0.2Q = 8 + 0.1Q

Which gives Q=40 or 4000 in hundreds of units.

Now that we integrated the noise into our considerations,  the equilibrium rental exceeds the social equilibrium rental by 10 (50-40) (1000 in hundreds of units)

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Step-by-step explanation:

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<em></em>

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Check the picture below.

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