Question

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 unit 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?

Answer 1

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)