If
C(x) = 13000 + 500x − 3.2x2 + 0.004x3
is the cost function and
p(x) = 2900 − 8x
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
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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 (MR) is the derivative of the demand function p(x), and the marginal cost (MC) is the derivative of the cost function C(x).
To find MR, we take the derivative of the demand function p(x):
MR = dp(x)/dx
MR = d(2900 - 8x)/dx
MR = -8
To find MC, we take the derivative of the cost function C(x):
MC = dC(x)/dx
MC = d(13000 + 500x - 3.2x^2 + 0.004x^3)/dx
MC = 500 - 6.4x + 0.012x^2
Now, we set MR equal to MC and solve for x:
-8 = 500 - 6.4x + 0.012x^2
0 = 8 + 6.4x - 0.012x^2
0.012x^2 - 6.4x - 8 = 0
We can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values for a, b, and c:
x = (-(-6.4) ± √((-6.4)^2 - 4(0.012)(-8))) / (2*0.012)
x = (6.4 ± √(40.96 + 0.384)) / 0.024
x = (6.4 ± √41.344) / 0.024
Now we can solve for x.
To find the production level that will maximize profit, we need to find the level of production where the marginal revenue equals the marginal cost.
First, let's find the total revenue function, TR(x), which is given by the product of the quantity produced (x) and the price function (p(x)):
TR(x) = x * p(x)
Substituting the given demand function p(x) = 2900 - 8x:
TR(x) = x * (2900 - 8x)
Next, let's find the marginal revenue function, MR(x), which is the derivative of the total revenue function with respect to x:
MR(x) = d(TR(x)) / dx
Differentiating the total revenue function:
MR(x) = d(x * (2900 - 8x)) / dx
= 2900 - 16x
Now let's find the cost function, C(x):
C(x) = 13000 + 500x - 3.2x^2 + 0.004x^3
The marginal cost function, MC(x), is the derivative of the cost function with respect to x:
MC(x) = d(C(x)) / dx
Differentiating the cost function:
MC(x) = d(13000 + 500x - 3.2x^2 + 0.004x^3) / dx
= 500 - 6.4x + 0.012x^2
To maximize profit, we need to find the production level where the marginal revenue equals the marginal cost. In other words, we need to find the value of x where MR(x) = MC(x):
2900 - 16x = 500 - 6.4x + 0.012x^2
This is a quadratic equation. Rearranging and setting the equation equal to zero, we get:
0.012x^2 - 9.6x + 2400 = 0
To solve for x, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
where a = 0.012, b = -9.6, and c = 2400.
Using the quadratic formula, we can find the two possible values of x. The value of x that maximizes profit will be the one that is feasible in the given context (e.g., it falls within the domain of the problem, or it gives a positive value for production).