Answer:
(a) The number of units that will give the maximum profit is 2,000 units.
(b) The maximum possible profit is $274,000.
Step-by-step explanation:
(a) How many units will give the maximum profit?
From the attached photo in the question, we are given;
R(x) = 324x - 0.0495x^2 ………………....................................……….. (1)
C(x) = 8000 + 162x - 0.099x^2 + 0.00003x^3 …………………….. (2)
In economics, profit function, P(x), is obtained as follows:
P(x) = R(x) - C(x) ………………………. (3)
Substituting the equations (1) and (2) for R(x) and C(x) respectively into equation (3) and then simplify, we have:
P = 324x - 0.0495x^2 - (8000 + 162x - 0.099x^2 + 0.00003x^3)
P = 324x - 0.0495x^2 - 8000 - 162x + 0.099x^2 - 0.00003x^3
P = -8000 + 162x + 0.0495x^2 - 0.00003x^3 ………………….. (4)
Taking a derivative of equation (4) with respect to x and equating to 0, we have:
162 + 0.099x - 0.00009x^2 = 0 .................... (5)
Using the almighty formula as follows:
x = (-b +/- (b^2 - 4ac)^0.5) / 2a ………………. (6)
Where, from equation (5);
a = - 0.00009
b = 0.099
c = 162
Substituting the values into equation (6), we have:
x = (-0.099 +/- (0.099^2 - (4 * (- 0.00009) * 162))^0.5) / (2 * - 0.00009)
x = (-0.099 +/- 0.261) / -0.00018
x = (-0.099 + 0.261) / -0.00018, or x = (-0.099 - 0.261) / -0.00018
x = -900, or x = 2000
Since negative number does not have economic value in economics, the number of units that will give the maximum profit is therefore 2,000 units.
(b) What is the maximum possible profit?
From part a, x = 2000
Substituting x = 2000 into equation (4) and then solve, we have:
P(x) = -8000 + (162 * 2000) + (0.0495 * 2000^2) - (0.00003 * 2000^3)
P(x) = 274,000
Therefore, the maximum possible profit is $274,000.
