Question

a product is introduced to the market. the weekly profit (in dollars) of that product decays exponentially as function of the price that is charged (in dollars) and is given by $p(x)=85000cdot e^{-0.04cdot x}$. suppose the price in dollars of that product, $x(t)$, changes over time $t$ (in weeks) as given by $x(t)=52 + 0.61cdot t^{2}$. find the rate that profit changes as a function of time, $p(t)$ dollars/week. how fast is profit changing with respect to time 3 weeks after the introduction.

Explanation:

Step1: Apply the chain - rule

The chain - rule states that if $P(x)$ is a function of $x$ and $x(t)$ is a function of $t$, then $P^{\prime}(t)=\frac{dP}{dx}\cdot\frac{dx}{dt}$. First, find $\frac{dP}{dx}$ for $P(x) = 85000e^{-0.04x}$. Using the derivative formula for $y = ae^{bx}$ ($y^\prime=abe^{bx}$), we have $\frac{dP}{dx}=85000\times(- 0.04)e^{-0.04x}=-3400e^{-0.04x}$.

Step2: Find $\frac{dx}{dt}$

Given $x(t)=52 + 0.61t^{2}$, using the power - rule for differentiation ($\frac{d}{dt}(at^{n})=nat^{n - 1}$), we get $\frac{dx}{dt}=1.22t$.

Step3: Calculate $P^{\prime}(t)$

Substitute $\frac{dP}{dx}$ and $\frac{dx}{dt}$ into the chain - rule formula: $P^{\prime}(t)=\frac{dP}{dx}\cdot\frac{dx}{dt}=(-3400e^{-0.04x})\cdot(1.22t)$. Substitute $x = 52+0.61t^{2}$ into the equation: $P^{\prime}(t)=-3400\times1.22t\cdot e^{-0.04(52 + 0.61t^{2})}=-4148t e^{-2.08-0.0244t^{2}}$.

Step4: Evaluate $P^{\prime}(t)$ at $t = 3$

Substitute $t = 3$ into $P^{\prime}(t)$:
\[

$$\begin{align*} P^{\prime}(3)&=-4148\times3\times e^{-2.08-0.0244\times3^{2}}\\ &=-12444\times e^{-2.08 - 0.2196}\\ &=-12444\times e^{-2.2996}\\ &\approx-12444\times0.1018\\ &=- 1266.7992\approx - 1266.80 \end{align*}$$

\]

Answer:

$-1266.80$