A manufacturer of small chain saws has determined:
R(x)=xp(x)=x(200-x/30)=200x-x^2/30
C(x)=7200+60x

Where x is the number of saws that can be sold at price $p per saw and C(x) is the total cost (in dollars) of producing x saws.
a. Find the marginal cost
b. Find the marginal revenue
c. What is the minimum number of saws sold for the company to break even?
d. Find the profit function and write its equation.
e. Find the marginal profit.
f. Find the number of saws produced that generate maximum profit.
g. What is the average cost per saw at maximum profit?
h. What is the maximum profit?

The problem is I have the worst time with derivatives and since I do not know how to simplify to get the equation for P(x)=R(x)-C(x), it makes completing a-h impossible since I do not know the correct equation. In previous problems I have done marginal profit. Is marginal cost and marginal revenue the same equation?

Solution Preview

a. Find the marginal cost
Since C(x)=7200+60x, C'(x)=60. Hence marginal cost=60.

b. Find the marginal revenue
Since R(x)=xp(x)=x(200-x/30)=200x-x^2/30, R'(x)=200-x/15. Hence marginal revenue = 200-x/15

c. What is the minimum number of saws sold for ...

Solution Summary

The revenues and cost calculations are examined in the solution.

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