If C(x) = 16000 + 600x − 1.8x2 + 0.004x3 is the cost function and p(x) = 4200 − 6x is the demand function, find the production level that will maximize profit. (Hint: If the profit is maximized, then the marginal revenue equals the marginal cost.)

Quantity that will maximize profit=1000

Step-by-step explanation:

Assume quantity=x

Revenue=price*quantity

=(4200-6x)x

=4200x-6x^2

Marginal revenue(MR) =4200-12x

Cost(x)= 16000 + 600x − 1.8x2 + 0.004x3

Marginal cost(MC) =600-3.6x+0.012x^2

Marginal cost=Marginal revenue

600-3.6x+0.012x^2=4200-12x

600-3.6x+0.012x^2-4200+12x=0

0.012x^2-8.4x-3600=0

Solve the quadratic equation using

x= -b +or- √b^2-4ac/2a

a=0.012

b=-8.4

c=-3600

x=-(-8.4) +or- √(-8.4)^2- (4)(0.012)(-3600) / (2)(0.012)

= 8.4 +or- √70.56-(-172.8) / 0.024

= 8.4 +or- √70.56+172.8 / 0.024

= 8.4 +or- √243.36 / 0.024

= 8.4 +or- 15.6/0.024

= 8.4/0.024 +15.6/0.024

= 350+650

x=1000

OR

= 8.4/0.024 -15.6/0.024

= 350 - 650

= -300

x=1000 or -300

Quantity that maximises profits can not be negative

So, quantity that maximises profits=1000

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.