The cost function for a product is \( C(x)=0.6 x^{2}+110 x+100 \) Find average cost over \( [0,500] \) Answer:

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 \( b = 500 \). Therefore, we need to compute: \[ A = \frac{1}{500 - 0} \int_0^{500} C(x) \, dx = \frac{1}{500} \int_0^{500} (0.6 x^2 + 110 x + 100) \, dx \] Now, we will calculate the integral: \[ \int (0.6 x^2 + 110 x + 100) \, dx = 0.6 \cdot \frac{x^3}{3} + 110 \cdot \frac{x^2}{2} + 100x + C \] Calculating the definite integral from \(0\) to \(500\): \[ \int_0^{500} (0.6 x^2 + 110 x + 100) \, dx = \left[ 0.6 \cdot \frac{x^3}{3} + 110 \cdot \frac{x^2}{2} + 100x \right]_0^{500} \] Calculating at \(x = 500\): \[ = 0.6 \cdot \frac{500^3}{3} + 110 \cdot \frac{500^2}{2} + 100 \cdot 500 \] Calculating each term: 1. \( 0.6 \cdot \frac{500^3}{3} = 0.6 \cdot \frac{125000000}{3} = 0.6 \cdot 41666666.67 \approx 25000000 \) 2. \( 110 \cdot \frac{500^2}{2} = 110 \cdot \frac{250000}{2} = 110 \cdot 125000 = 13750000 \) 3. \( 100 \cdot 500 = 50000 \) Now, summing these values: \[ 25000000 + 13750000 + 50000 = 38700000 \] Now, we can find the average cost: \[ A = \frac{1}{500} \cdot 38700000 = 77400 \] Thus, the average cost over the interval \([0, 500]\) is: \[ \boxed{77400} \]