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To find the cost of producing 10 incremental units after 15 units have been produced, we need to determine the total cost of increasing production from 15 units to 25 units. This can be accomplished by integrating the marginal cost function from 15 to 25.
The marginal cost [tex]$C'(x)$[/tex] is given by:
[tex]\[ C'(x) = 85 + \frac{375}{x^2} \][/tex]
1. Set up the integral:
To find the total cost of producing the additional units, we integrate the marginal cost function from [tex]\( x = 15 \)[/tex] to [tex]\( x = 25 \)[/tex]:
[tex]\[ \int_{15}^{25} \left( 85 + \frac{375}{x^2} \right) dx \][/tex]
2. Integrate the function:
We need to find the antiderivative of the integrand. The integral can be broken into two parts:
[tex]\[ \int_{15}^{25} 85 \, dx + \int_{15}^{25} \frac{375}{x^2} \, dx \][/tex]
a. For the first part:
[tex]\[ \int_{15}^{25} 85 \, dx \][/tex]
The antiderivative of [tex]\( 85 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 85x \)[/tex].
Evaluate this from [tex]\( 15 \)[/tex] to [tex]\( 25 \)[/tex]:
[tex]\[ 85x \bigg|_{15}^{25} = 85(25) - 85(15) \][/tex]
b. For the second part:
[tex]\[ \int_{15}^{25} \frac{375}{x^2} \, dx \][/tex]
Recall that [tex]\( \frac{1}{x^2} \)[/tex] is the same as [tex]\( x^{-2} \)[/tex]. The antiderivative of [tex]\( \frac{375}{x^2} \)[/tex] is [tex]\( -\frac{375}{x} \)[/tex].
Evaluate this from [tex]\( 15 \)[/tex] to [tex]\( 25 \)[/tex]:
[tex]\[ -\frac{375}{x} \bigg|_{15}^{25} = -\frac{375}{25} - \left( -\frac{375}{15} \right) \][/tex]
3. Calculate the definite integrals:
a. For the first integral:
[tex]\[ 85(25) - 85(15) = 2125 - 1275 = 850 \][/tex]
b. For the second integral:
[tex]\[ -\frac{375}{25} + \frac{375}{15} = -15 + 25 = 10 \][/tex]
4. Sum the results of the integrals:
[tex]\[ 850 + 10 = 860 \][/tex]
Therefore, the cost of producing the additional 10 units after 15 units have been produced is $860.
The marginal cost [tex]$C'(x)$[/tex] is given by:
[tex]\[ C'(x) = 85 + \frac{375}{x^2} \][/tex]
1. Set up the integral:
To find the total cost of producing the additional units, we integrate the marginal cost function from [tex]\( x = 15 \)[/tex] to [tex]\( x = 25 \)[/tex]:
[tex]\[ \int_{15}^{25} \left( 85 + \frac{375}{x^2} \right) dx \][/tex]
2. Integrate the function:
We need to find the antiderivative of the integrand. The integral can be broken into two parts:
[tex]\[ \int_{15}^{25} 85 \, dx + \int_{15}^{25} \frac{375}{x^2} \, dx \][/tex]
a. For the first part:
[tex]\[ \int_{15}^{25} 85 \, dx \][/tex]
The antiderivative of [tex]\( 85 \)[/tex] with respect to [tex]\( x \)[/tex] is [tex]\( 85x \)[/tex].
Evaluate this from [tex]\( 15 \)[/tex] to [tex]\( 25 \)[/tex]:
[tex]\[ 85x \bigg|_{15}^{25} = 85(25) - 85(15) \][/tex]
b. For the second part:
[tex]\[ \int_{15}^{25} \frac{375}{x^2} \, dx \][/tex]
Recall that [tex]\( \frac{1}{x^2} \)[/tex] is the same as [tex]\( x^{-2} \)[/tex]. The antiderivative of [tex]\( \frac{375}{x^2} \)[/tex] is [tex]\( -\frac{375}{x} \)[/tex].
Evaluate this from [tex]\( 15 \)[/tex] to [tex]\( 25 \)[/tex]:
[tex]\[ -\frac{375}{x} \bigg|_{15}^{25} = -\frac{375}{25} - \left( -\frac{375}{15} \right) \][/tex]
3. Calculate the definite integrals:
a. For the first integral:
[tex]\[ 85(25) - 85(15) = 2125 - 1275 = 850 \][/tex]
b. For the second integral:
[tex]\[ -\frac{375}{25} + \frac{375}{15} = -15 + 25 = 10 \][/tex]
4. Sum the results of the integrals:
[tex]\[ 850 + 10 = 860 \][/tex]
Therefore, the cost of producing the additional 10 units after 15 units have been produced is $860.
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