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 Bryant Hanson. in the United States

Mar 12,2025

To find the local maximum for the revenue function \( R(x) = 1620x - 0.15x^3 \), we need to take the first derivative and set it to zero. The first derivative is: \[ R'(x) = 1620 - 0.45x^2 \] Setting \( R'(x) = 0 \): \[ 1620 - 0.45x^2 = 0 \] Solving for \( x \): \[ 0.45x^2 = 1620 \implies x^2 = \frac{1620}{0.45} \implies x^2 = 3600 \implies x = 60 \] Next, we can check the second derivative to confirm it’s a maximum: \[ R''(x) = -0.90x \] At \( x = 60 \): \[ R''(60) = -0.90 \times 60 = -54 < 0 \] Since the second derivative is negative, there is a local maximum at \( x = 60 \). Therefore, the answer for part (A) is: The local maximum is at \( x = 60 \). As for part (B), stating “There is no local maximum” would not be correct, as we have established there is indeed a local maximum at \( x = 60 \).