A factory produces equipment whose wholesale price is $50 each. It is stated that the role of income is I (x) = 84 x-1000 where x is the number of units produc

Question

A factory produces equipment whose wholesale price is $50 each.

 

It is stated that the role of income is I (x) = 84 x-1000 where x is the number of units produced.

 

Similarly, the costs associated with that production can be calculated using C (x) = 0.4x2 - 60 x + 8000

 

 

Find: The number of units to produce for minimal cost. The function of utilities P (x) which is obtained by subtracting the cost of admission.

 

The range of values of x, in which there is profit, i.e. when P(x) > 0.

 

The number of units that is necessary to produce for the higher profit and the value of the same

Answer 1

C(x) = .4x^2 -60x +2000

C' = .8x -60 = 0

x=60/.8 = 75 = cost minimizing output level

min C = .4(75)^2 -60(75) + 2000

2250 - 4500 + 2000 = -250

minimum cost is negative = a $250 profit

I(x) = 84x -1000

I'(x) = 84

max profit generating output is when I'(x) = C'(x)

84 = .8x -60

.8x = 144

x= 144/.8

x=180

I(180) -C(180) = 84(180) - .4(180)^2 +60(180) -8000

= 144(180) -8000 - 12960

= 25920 - 20960 = $4,960 = max profit when x=180

profit is when P(x)>0 when I(x)>C(x)

84x -1000 > .4x^2 -60x + 8000

0 > .4x^2 -144x +9000

.4x^2 -144x + 9000 > 0

breakeven output level

x = 144/.8 + or - (1/.8)sqr(144^2 -1.6(9000))

x = 180 + or - 1.25sqr(20736 - 14400)

= 180 + or - 1.25sqr6336

= 180 + or - 1.25(79.6) = 180+99.5

x = 279.5, or 80.5

80.5 < x< 279.5 is the domain for a positive profit

the midpoint output level is the profit maximizing output=180

I(75) =84(75)-1000 = 6300-1000= 5300

max profit = 5300--240 = $5,540