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A small co ffee company roasts coff ee beans in its shop. The unroasted beans cost the com- pany 200 cents per pound. The MARGINAL cost of roasting coff ee beans is 150 - 10q + q^2 cents per pound when q pounds are roasted. The smell of roasting beans imposes costs on the company's neighbors. The total amount that neighbors would be willing to pay to have the shop stop roasting altogether is 5q^2; where q is the number of pounds being roasted. The company sells its output in a competitive market at 450 cents per pound. What is the socially efficient amount of co ffee for the company to roast?

Sagot :

Answer:

q = 10 pounds = socially efficient amount of coffee for the company to roast.

Explanation:

Data Given:

Cost of Unroasted beans = 200 cents/pound.

Marginal Cost of roasting coffee beans = [tex]q^{2}[/tex]-10[tex]q^{}[/tex] + 150

Cost neighbors willing to pay to stop shop operations = 5[tex]q^{2}[/tex]

Output selling price = 450 cents/pound

Required:

Amount of coffee to roast = ?

Solution:

As we know from the problem statement that it costs the company 200 cents/pound for the procurement of raw beans which are here termed as unroasted beans. Let's say it is the marginal cost of procuring.

Moreover, we know that marginal cost of roasted beans = [tex]q^{2}[/tex]-10[tex]q^{}[/tex] + 150.

which is in the form of quadratic equation and will be solved for q to know the required answer.

Let's suppose X = marginal cost of unroasted beans.

Y = marginal cost of roasted beans.

MPC = Marginal Private cost

In order to calculate the marginal private cost, we need to add X+Y.

MPC = 200 + [tex]q^{2}[/tex]-10[tex]q^{}[/tex] + 150

MPC = [tex]q^{2}[/tex]-10[tex]q^{}[/tex] + 350

Now,

The total social cost which the neighbors are willing to pay = 5[tex]q^{2}[/tex]

In order to calculate marginal social cost, we need to differentiate the above equation.

MSC = d/dq5[tex]q^{2}[/tex] = 10q

Finally,

Marginal Benefit = 450 cents/pound

For socially efficient amount = q =

MPC + MSC = Marginal Benefit

[tex]q^{2}[/tex]-10[tex]q^{}[/tex] + 350 + 10[tex]q^{}[/tex] = 450

[tex]q^{2}[/tex]-10[tex]q^{}[/tex] + 10[tex]q^{}[/tex] = 450-350

Solving for q,

[tex]q^{2}[/tex] = 100

taking square root on both sides,

q = +/-10.

Hence,

q = 10 pounds = socially efficient amount of coffee for the company to roast.

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