The marketing research department of a computer company used a large city to test market the firm's new laptop. The department found the relationship between price \( p \) (dollars per unit) and the demand \( x \) (units per week) was given approximately by the following equation. \[ p=1620-0.15 x^{2} \quad 0

Ask by Macdonald Norton. in the United States

Mar 12,2025

To find the local maximum and minimum of the revenue function \( R(x) = 1620x - 0.15x^3 \), we start by taking the first derivative \( R'(x) \) and setting it equal to zero to find critical points. The first derivative is: \[ R'(x) = 1620 - 0.45x^2 \] Setting \( R'(x) = 0 \): \[ 1620 - 0.45x^2 = 0 \] \[ 0.45x^2 = 1620 \] \[ x^2 = \frac{1620}{0.45} = 3600 \] \[ x = 60 \] To determine if this point is a maximum or minimum, we can check the second derivative: \[ R''(x) = -0.90x \] Evaluating \( R''(60) \): \[ R''(60) = -0.90 \times 60 = -54 \] Since \( R''(60) < 0 \), this indicates a local maximum at \( x = 60 \). Thus, for your queries: A. The local maximum is at \( x = 60 \). Next, since the revenue function is a cubic polynomial that opens downwards due to the negative leading coefficient, and it has no lower bounds within the interval \( 0 < x < 80 \), we can conclude: B. There is no local minimum. Final answers: A. The local maximum is at \( x = 60 \). B. There is no local minimum.