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

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To solve the problem, we will analyze the revenue function $$ R(x) = 1620x - 0.15x^3 $$ to find local minima and determine the concavity of the graph. ### Part A: Finding Local Minimum 1. **Find the first derivative of $$ R(x) $$**: $$ R'(x) = \frac{d}{dx}(1620x - 0.15x^3) = 1620 - 0.45x^2 $$ 2. **Set the first derivative equal to zero to find critical points**: $$ 1620 - 0.45x^2 = 0 $$ Rearranging gives: $$ 0.45x^2 = 1620 \quad \Rightarrow \quad x^2 = \frac{1620}{0.45} = 3600 \quad \Rightarrow \quad x = \sqrt{3600} = 60 $$ 3. **Determine if this critical point is a local minimum or maximum by using the second derivative test**: $$ R''(x) = \frac{d^2}{dx^2}(1620x - 0.15x^3) = -0.90x $$ Evaluate $$ R''(60) $$: $$ R''(60) = -0.90 \times 60 = -54 $$ Since $$ R''(60) < 0 $$, this indicates that $$ x = 60 $$ is a local maximum, not a minimum. Thus, there are no local minima in the interval $$ 0 < x < 80 $$. ### Part B: Concavity of the Revenue Function 1. **Determine the intervals of concavity using the second derivative**: - The second derivative is $$ R''(x) = -0.90x $$. - Set $$ R''(x) = 0 $$ to find inflection points: $$ -0.90x = 0 \quad \Rightarrow \quad x = 0 $$ - Analyze the sign of $$ R''(x) $$: - For $$ x > 0 $$, $$ R''(x) < 0 $$ (the graph is concave downward). - For $$ x < 0 $$, $$ R''(x) > 0 $$ (not applicable since $$ x > 0 $$). Since the only relevant interval is $$ 0 < x < 80 $$ and $$ R''(x) < 0 $$ throughout this interval, the graph is never concave upward. ### Final Answers A. The local minimum/a is/are at $$ x= $$ (There is no local minimum). B. The graph is never concave upward. A. There is no local minimum. B. The graph is never concave upward.

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