K(x)=0.02x^2+72 A. Set up the equation for the average costs B. Calculate the average costs C.set the marginal cost function to D. Calculate the marginal costs for 50 products E. At what quantity does one reach the operating optimum

\text{Average Cost}(50) = \frac{0.02(50)^2 + 72}{50} = \frac{0.02 \cdot 2500 + 72}{50} = \frac{50 + 72}{50} = 2.44 [Step-by-Step] \text{Average Cost}(50) = \frac{0.02(50)^2 + 72}{50} = \frac{0.02 \cdot 2500 + 72}{50} = \frac{50 + 72}{50} = 2.44

C. Set the marginal cost function necessarily [Solution] \text{Marginal Cost}(x) = \frac{dK(x)}{dx} = \frac{d}{dx}(0.02x^2 + 72) = 0.04x [Step-by-Step] \text{Marginal Cost}(x) = \frac{dK(x)}{dx} = \frac{d}{dx}(0.02x^2 + 72) = 0.04x

D. Calculate the marginal costs for 50 products necessarily [Solution] \text{Marginal Cost}(50) = 0.04 \cdot 50 = 2 [Step-by-Step] \text{Marginal Cost}(50) = 0.04 \cdot 50 = 2

E. At what quantity does one reach the operating optimum necessarily [Solution] \text{Operating Optimum is reached when Marginal Cost equals Average Cost} \text{Set } 0.04x = \frac{0.02x^2 + 72}{x} 0.04x = 0.02x + \frac{72}{x} 0.02x^2 = 72 x^2 = 3600 x = 60 [Step-by-Step] \text{Operating Optimum is reached when Marginal Cost equals Average Cost} \text{Set } 0.04x = \frac{0.02x^2 + 72}{x} 0.04x = 0.02x + \frac{72}{x} 0.02x^2 = 72 x^2 = 3600 x = 60