The total cost m millions of dollars of producing r thousand

The total cost m millions of dollars, of producing r thousand units of an item is C(x) = x^2 + 6x + 3 The revenue (in millions of dollars) from selling x thousand units of the item is R(x) = 10x Profit is revenue minus cost For what value of r does the firm make profit? Break even? Lose money? Find the value of x that maximizes profit and the corresponding profit

Solution

Total cost, C(x)= x²+6x+3

Revenue, R(x) = 10x

Profit,P(x) = R(x) - C(x)

For profit, P(x) > 0

              R(x) - C(x) >0

R(x) > C(x)

10x>x²+6x+3

x²-4x+3<0

(x-3)(x-1)<0

1<x<3

Lose money is in the case when profit <0

R(x)-C(x)<0

10x-x²-6x-3<0

(x-3)(x-1)>0

x>3   and x<1

Break even

Total profit at break even point is 0

R(x) - C(x)=0

R(x)=C(x)

x²+6x+3=10x

x²-4x+3=0

(x-3)(x-1)=0

x=1 and x=3

b. Profit is maximum at maximum value of x

P(x)=10x-x²-6x-3 =-x²+4x -3

And this is a quadratic equation in which x is maximum at the vertex

x=-b/2a   = -4/-2= 2

P(2)= -(2)²+4(2) -3

      = 1