A dairy is planning to introduce and promote a new line of organic ice cream. After test marketing the new line in a large city, the marketing research department found that the demand in that city is given approximately by the following equation, where \( x \) thousand quarts were sold per week at a price of \( \$ p \) each, and whose revenue function is given as \( R(x)=x p \). \( p=16 e^{-x} \quad 0

Ask by Riley Vega. in the United States

Mar 12,2025

**(A) Finding the Local Extrema** The revenue function is given by \[ R(x)=16xe^{-x}. \] To find the critical points, we first compute the derivative \(R'(x)\) using the product rule. Recall that the derivative of \(e^{-x}\) is \(-e^{-x}\). Thus, \[ R'(x)=\frac{d}{dx}\big(16x \cdot e^{-x}\big)=16\left(e^{-x} + x(-e^{-x})\right)=16\left(e^{-x}-xe^{-x}\right)=16e^{-x}(1-x). \] We set the derivative equal to zero: \[ 16e^{-x}(1-x)=0. \] Since \(16e^{-x}\) is never zero, we require \[ 1-x=0 \quad \Rightarrow \quad x=1. \] To classify the critical point, we compute the second derivative \(R''(x)\). Differentiate \(R'(x)=16e^{-x}(1-x)\) using the product rule: \[ R''(x)=16\left(\frac{d}{dx}\big(e^{-x}(1-x)\big)\right). \] Let \(u=e^{-x}\) and \(v=1-x\). Then, \(u'= -e^{-x}\) and \(v'=-1\). So, \[ \frac{d}{dx}\big(e^{-x}(1-x)\big)=u'v+uv'=-e^{-x}(1-x)+e^{-x}(-1)=-e^{-x}(1-x)-e^{-x}. \] Simplify: \[ -e^{-x}(1-x)-e^{-x} = -e^{-x}((1-x)+1)=-e^{-x}(2-x). \] Thus, \[ R''(x)=-16e^{-x}(2-x). \] Evaluate \(R''(x)\) at \(x=1\): \[ R''(1)=-16e^{-1}(2-1)=-16e^{-1}. \] Since \(-16e^{-1}<0\), the function is concave downward at \(x=1\) which indicates a local maximum at this point. The revenue at \(x=1\) is \[ R(1)=16(1)e^{-1}=\frac{16}{e}. \] Thus, the local maximum is at: \[ \boxed{x=1}. \] **(B) Determining Concavity** The second derivative is \[ R''(x)=-16e^{-x}(2-x). \] Since \( -16e^{-x} < 0 \) for all \(x\) (because \(e^{-x}>0\) for all \(x\)), the sign of \(R''(x)\) depends on the factor \((2-x)\). - When \((2-x) > 0\), i.e., \(x < 2\): \[ R''(x) < 0, \] so the graph is concave downward. - When \((2-x) < 0\), i.e., \(x > 2\): \[ R''(x) > 0, \] so the graph is concave upward. Given the domain \(0