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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 Maxwell Osborne. in the United States
Mar 12,2025

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A. There is no local minimum. B. The graph is never concave upward.

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To find the local minimum for the revenue function \( R(x) = 1620x - 0.15x^3 \), we can start by taking the derivative and setting it to zero to find critical points. The first derivative is \( R'(x) = 1620 - 0.45x^2 \). Setting this equal to zero gives us \( 1620 = 0.45x^2 \), leading to \( x^2 = 3600 \), so \( x = 60 \) is a critical point within the viable range (0 < x < 80). Now for concavity, the second derivative \( R''(x) = -0.9x \) tells us the concavity of the graph of the revenue function. This expression is positive when \( x < 0 \) and negative when \( x > 0 \). Consequently, the graph is never concave upward as \( x \) must be positive within the given range. So, for your questions: A. The local minimum is at \( x = 60 \). B. The graph is never concave upward.

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