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 ) = 95000 ⋅ e − 0.05 ⋅ x Suppose the price in dollars of that product, x ( t ) , changes over time t (in weeks) as given by x ( t ) = 53 + 0.95 ⋅ 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 7 weeks after the introduction. dollars/week

Answer 1

Answer:

1). $P'(t) = (-9025t).e^{-0.05(53+0.95t^2)}$

2). (-435.36) dollars per week

Step-by-step explanation:

Weekly price decay of the product is represented by the function,

P(x) = $95000.e^{-0.05x}$

And the price of the product changes over the period of 't' weeks is represented by,

x(t) = $53+0.95t^2$

Function representing the rate of change in the profit with respect to the time will be represented by,

1). P'(t) = $\frac{dP}{dx}.\frac{dx}{dt}$

Since, P(x) = $95000.e^{-0.05x}$

P'(x) = $95000\times (-0.05).e^{-0.05x}$

       = $(-4750).e^{-0.05x}$

Since, x(t) = 53 + 0.95t²

x'(t) = 1.9t

$\frac{dP}{dx}.\frac{dx}{dt}=(-4750).e^{-0.05x}\times (1.9t)$

By substituting x = 53 + 0.95t²

$\frac{dP}{dx}.\frac{dx}{dt}=(-4750).e^{-0.05(53+0.95t^2)}\times (1.9t)$

   P'(t) = $(-9025t).e^{-0.05(53+0.95t^2)}$

2). For t = 7 weeks,

P'(7) = $(-9025\times 7).e^{-0.05(53+0.95(7)^2)}$

       = $(-63175).e^{-4.9775}$

       = (-63175)(0.006891)

       = (-435.356) dollars per week

       ≈ (-435.36) dollars per week