Applied Example: The profit function of Acrosonic Company is given by \( P(x)=-0.02 x^{2}+300 x-200,000 \) dollars, where \( x \) is the number of Acrosonic model \( F \) loudspeakers systems produced. Find the extrema points of the \( P(x) \) function.

To find the extrema points of the profit function \( P(x)=-0.02 x^{2}+300 x-200,000 \), we first need to calculate the derivative of the function and set it equal to zero. The derivative \( P'(x) \) gives us: \[ P'(x) = -0.04x + 300. \] Setting \( P'(x) = 0 \) leads us to: \[ -0.04x + 300 = 0 \] \[ 0.04x = 300 \] \[ x = 7500. \] Next, to determine whether this point is a maximum or minimum, we can use the second derivative test. The second derivative \( P''(x) \) is: \[ P''(x) = -0.04. \] Since \( P''(x) < 0 \), this indicates that \( P(x) \) has a local maximum at \( x = 7500 \). Thus, the extremum point occurs when 7500 loudspeakers are produced, which yields the maximum profit. Now, let’s calculate the maximum profit by substituting \( x = 7500 \) back into \( P(x) \): \[ P(7500) = -0.02(7500)^{2}+300(7500)-200,000. \] Calculating that gives: \[ P(7500) = -0.02(56250000) + 225000 - 200000 = -1125000 + 225000 - 200000 = 112500. \] Therefore, the maximum profit occurs when 7500 loudspeakers are produced, resulting in a profit of $112,500.