The cost function for producingq items is given by
C(q)=500+15q-250e^-0.04q
(a) Find the marginal cost when q = 0 and when q = 100 and interpret your results.
(b) Determine the number of items produced so that the marginal cost is less than $20
The cost function for producingq items is given by
C(q)=500+15q-250e^-0.04q
(a) Find the marginal cost when q = 0 and when q = 100 and interpret your results.
(b) Determine the number of items produced so that the marginal cost is less than $20
(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.