**Q: Word problem, here it is:
**

**(a) The marginal cost function for Janes Coffee Co is MC(q) = 40 – 2q, where q is the number of tons of coffee produced. Fixed costs are $250. The marginal revenue function is MR(q) = 100 – 4q. What is the value of the profit function P(q) when q = 20?
**

**(b) The average cost when q=20 is?**

**(c) The change in revenue if sales increase from 10 to 15 tons is?**

A: “Marginal” means cost per unit, otherwise known as the derivative in the calculus world. Fixed costs represent the cost at time 0 (or 0 units), in other words… the “+ C” part. Therefore, we know that:

(a)

C ‘ (q) = 40 – 2q

and the “+ C” when we integrate is 250.

R ‘ (q) = 100 – 4q

and the “+ C” when we integrate will be 0 (because there is no revenue when nothing is produced)

So, now let’s integrate and find C(q) and R(q).

C(q) = ∫40 – 2q dq = 40q – q² + C

C(q) = 40q – q² + 250

R(q) = ∫100 – 4q dq = 100q – 2q² + C

R(q) = 100q – 2q²

So far so good? We also know that Profit = Revenue – Cost

P(q) = R(q) – C(q)

P(q) = (100q – 2q²) – (40q – q² + 250)

Simplify now:

P(q) = 100q – 2q² – 40q + q² – 250

P(q) = -q² + 60q – 250

And, to finally answer the question:

P(20) = -(20)² + 60(20) – 250 = -400 + 1200 – 250 = 550

(b) To find the average cost, you take C(q) / q…..

Average Cost: C(q) / q = (40q – q² + 250) / q

Now, plug in q = 20

Average Cost = (40q – q² + 250) / q = (40(20) – (20)² + 250) / 20 = 650 / 20 = 65/2 = 32.5

(c) The change in revenue is the difference between the revenues when q is 10 and when q is 15:

R(q) = 100q – 2q²

R(10) = 100(10) – 2(10)² = 800

R(15) = 100(15) – 2(15)² = 1050

So, the change in revenue is 1050 – 800 = 250.