If
C(x) = 15000 + 600x − 0.6x^2 + 0.004x^3
is the cost function and
p(x) = 1800 − 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.)
To find the production level that will maximize profit, we need to find the point where the marginal revenue equals the marginal cost.
The marginal revenue is the derivative of the demand function p(x), and the marginal cost is the derivative of the cost function C(x).
First, let's find the derivative of the demand function p(x):
p'(x) = -6
Next, let's find the derivative of the cost function C(x):
C'(x) = 600 - 1.2x + 0.012x^2
Since the profit is maximized when the marginal revenue equals the marginal cost, we can set the two derivatives equal to each other and solve for x:
p'(x) = C'(x)
-6 = 600 - 1.2x + 0.012x^2
Now, let's solve this quadratic equation to find the values of x that satisfy the equation:
0.012x^2 - 1.2x + 606 = 0
We can use the quadratic formula to solve this equation:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In this equation, a = 0.012, b = -1.2, and c = 606.
Plugging in these values, we get:
x = (-(-1.2) ± sqrt((-1.2)^2 - 4 * 0.012 * 606)) / (2 * 0.012)
Simplifying further:
x = (1.2 ± sqrt(1.44 - 29.12)) / 0.024
x = (1.2 ± sqrt(-27.68)) / 0.024
Since the square root of a negative number is undefined in the real number system, this equation does not have any real solutions. Therefore, there is no production level that will maximize profit in this scenario.