(a) Find the marginal cost when q = 0 and when q = 100 and interpret your results.

Marginal cost is the first derivative of the cost function.

C(q) = 500 +15q -250e^(-0.04q)

d/dx a^u = ln(a) *a^u *du/dx

So,

C'(q) = 15 -250[ln(e) *e^(-0.04q) *(-0.04)]

C'(q) = 15 +(250)(0.04)e^(-0.04q)

C'(q) = 15 +10e^(-0.04q) ------------------------(i)

Hence,

C'(0) = 15 +10e^0 = 15 +10 = 25

C'(100) = 15 +10e^(-0.04*100) = 15 +0.183 = 15.183

The marginal cost is decreases as q increases. Marginal cost is inversely proportional to the demand q. ----------------answer.

(b) Determine the number of items produced so that the marginal cost is less than $20

C'(q) = 15 +10e^(-0.04q) ------------------------(i)

15 +10e^(-0.04q) < 20

10e^(-0.04q) < 5

e^(-0.04q) < 0.5

Take the natural logs of both sides,

(-0.04q)ln(e) < ln(0.5)

-0.04q < -0.693147

q < (-0.693147) /(-0.04)

Division by negative, so reverse the sense of the inequality,

q > 17.328

Therefore, so that the marginal cost is less than $20, produce 18 or more items. ---------------answer.