Someone analyzed data for 10 weeks of production and now estimates marginal cost for producing x units in a week (in thousands of dollars) to be C'(x)=50e^(-x/100). Given that fixed costs for a week are 3 million dollars, find the total cost of producing 100 units. I thought I was meant to find the antiderivative of this function and then plug in the two other values to find the total cost but when I did that I was wrong and now I'm confused.
too bad you didn't show your work, since you clearly want us to show ours.
C'(x) = 50e^(-x/100)
C(x) = -5000 e^(-x/100) + k
since C(0) = 3000 (the cost of producing 0 units is the fixed cost)
-5000 + k = 3000
k = 8000
C(x) = -5000e^(-x/100) + 8000
C(100) = -5000/e + 8000 = 6160
I see what I did wrong. I used the fixed cost as if it were k rather than plugging it in as a point. So the best way to find K is checking C(0) when having fixed cost?
To find the total cost of producing 100 units, we need to integrate the marginal cost function, C'(x), and then add the fixed costs.
The marginal cost function given is C'(x) = 50e^(-x/100).
The total cost function, C(x), can be found by integrating C'(x) with respect to x:
C(x) = ∫ C'(x) dx
Using the power rule of integration, we integrate C'(x) = 50e^(-x/100) as follows:
C(x) = ∫ 50e^(-x/100) dx
To integrate this function, we can use the substitution method. Let u = -x/100, which gives us du = -(1/100) dx.
Substituting these values into the integral, we get:
C(x) = ∫ 50e^u (-100 du)
C(x) = -5000 ∫ e^u du
Using the integral of e^u, which is e^u, our expression becomes:
C(x) = -5000e^u + C
Substituting u back in terms of x, we get:
C(x) = -5000e^(-x/100) + C
To find the constant of integration, C, we can use the given fixed costs of 3 million dollars when x = 0. So, let's substitute these values into the equation:
3,000,000 = -5000e^0 + C
Evaluating further, we have:
3,000,000 = -5000 + C
C = 3,005,000
Now that we have the constant of integration, we can rewrite our equation as:
C(x) = -5000e^(-x/100) + 3,005,000
To find the total cost of producing 100 units, substitute x = 100 into the total cost function:
C(100) = -5000e^(-100/100) + 3,005,000
C(100) = -5000e^(-1) + 3,005,000
Evaluating further, we have:
C(100) ≈ -5000(0.3679) + 3,005,000
C(100) ≈ -1839.5 + 3,005,000
C(100) ≈ 3,003,160.5
Therefore, the total cost of producing 100 units is approximately 3,003,160.5 dollars.
To find the total cost of producing 100 units, you indeed need to integrate the marginal cost function C'(x). However, there is an additional step you need to consider due to the fixed costs. Let's go through the process step by step.
The marginal cost function C'(x) represents the rate at which the cost changes with respect to the number of units produced in one week. In order to find the total cost function C(x) (rather than just the change in cost), you need to integrate C'(x) with respect to x:
C(x) = ∫C'(x) dx
Given that C'(x) = 50e^(-x/100), integrating C'(x) will give us the total cost function:
C(x) = ∫50e^(-x/100) dx
To solve this integral, we can use the power rule for integration:
∫e^ax dx = (1/a)e^ax + C
Applying this rule to our integral, we have:
C(x) = 50∫e^(-x/100) dx
= -50(100)e^(-x/100) + C
Now, let's consider the fixed costs. In a given week, the fixed costs are $3 million, regardless of the number of units produced. Thus, if no units are produced, the total cost is $3 million.
To incorporate the fixed costs into the total cost function, we need to add $3 million to our expression for C(x):
C(x) = -50(100)e^(-x/100) + 3,000,000
Finally, to find the total cost of producing 100 units, substitute x = 100 into the total cost function:
C(100) = -50(100)e^(-100/100) + 3,000,000
= -50(100)e^(-1) + 3,000,000
To evaluate this expression, we can use the approximation e^(-1) ≈ 0.3679:
C(100) ≈ -50(100)(0.3679) + 3,000,000
Evaluating further:
C(100) ≈ -18,395 + 3,000,000
≈ 2,981,605
Therefore, the total cost of producing 100 units is approximately $2,981,605.