A Brazilian manufacturer knows that if x (Hundred) products are demanded in a particular week.
(i) the total cost function in Brazilian real (R$000) is 14+3x and
(ii) the total revenue function (R$000) is 19x - 2x^2
a) Derive the total profit function
b) Find the profit break even points
c) Calculate the level of demand that maximises profits (ie) Maximum Profit point) and the amount of profit obtained
a) The profit function is given by the difference between the total revenue and the total cost:
Profit = Revenue - Cost
Profit = (19x - 2x^2) - (14 + 3x)
Simplifying this expression, we get:
Profit = 19x - 2x^2 - 14 - 3x
b) To find the profit break-even points, we need to find the values of x for which the profit is equal to zero. We set the profit function equal to zero and solve for x:
19x - 2x^2 - 14 - 3x = 0
-2x^2 + 16x - 14 = 0
Dividing through by -2, we get:
x^2 -8x + 7 = 0
This quadratic equation can be factored as:
(x - 1)(x - 7) = 0
Setting each factor equal to zero, we find two solutions:
x - 1 = 0 --> x = 1
x - 7 = 0 --> x = 7
Therefore, the profit break-even points occur when the demand is 1 hundred products or 7 hundred products.
c) To find the level of demand that maximizes profits, we can take the derivative of the profit function with respect to x and set it equal to zero.
dProfit/dx = 19 - 4x - 3 = 0
Simplifying this expression, we get:
-4x + 16 = 0
Solving for x, we find:
-4x = -16
x = 4
Therefore, the level of demand that maximizes profits is 4 hundred products.
To find the maximum profit, we substitute this value of x into the profit function:
Profit = 19(4) - 2(4^2) - 14 - 3(4)
Profit = 76 - 32 - 14 - 12
Profit = 18
So, the maximum profit obtained is R$18,000.