Hardy Curry
08/17/2024 · Primary School
The cost function for a product is \( C(x)=0.6 x^{2}+110 x+100 \). Find average cost over \( [0,500] \) Answer:
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Step-by-step Solution
To find the average cost over the interval \([0, 500]\), we first need to calculate the total cost over that interval and then divide by the length of the interval.
The average cost \( A \) over the interval \([a, b]\) is given by the formula:
\[
A = \frac{1}{b - a} \int_a^b C(x) \, dx
\]
In this case, \( a = 0 \) and \( b = 500 \). The cost function is given by:
\[
C(x) = 0.6x^2 + 110x + 100
\]
Now, we need to compute the integral:
\[
\int_0^{500} C(x) \, dx = \int_0^{500} (0.6x^2 + 110x + 100) \, dx
\]
We can calculate this integral term by term:
1. **Integrate \( 0.6x^2 \)**:
\[
\int 0.6x^2 \, dx = 0.6 \cdot \frac{x^3}{3} = 0.2x^3
\]
2. **Integrate \( 110x \)**:
\[
\int 110x \, dx = 110 \cdot \frac{x^2}{2} = 55x^2
\]
3. **Integrate \( 100 \)**:
\[
\int 100 \, dx = 100x
\]
Now, combining these results, we have:
\[
\int C(x) \, dx = 0.2x^3 + 55x^2 + 100x
\]
Next, we evaluate this integral from \( 0 \) to \( 500 \):
\[
\int_0^{500} C(x) \, dx = \left[ 0.2(500)^3 + 55(500)^2 + 100(500) \right] - \left[ 0.2(0)^3 + 55(0)^2 + 100(0) \right]
\]
Calculating each term:
1. \( 0.2(500)^3 = 0.2 \cdot 125000000 = 25000000 \)
2. \( 55(500)^2 = 55 \cdot 250000 = 13750000 \)
3. \( 100(500) = 50000 \)
Now, summing these values:
\[
\int_0^{500} C(x) \, dx = 25000000 + 13750000 + 50000 = 38700000
\]
Now, we can find the average cost:
\[
A = \frac{1}{500 - 0} \int_0^{500} C(x) \, dx = \frac{38700000}{500} = 77400
\]
Thus, the average cost over the interval \([0, 500]\) is:
\[
\boxed{77400}
\]
Quick Answer
The average cost over the interval \([0, 500]\) is \( 77400 \).
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