suppose that the demand function for a product is given by $d(p)=\frac{60000}{p}$ and that the price $p$ is…

suppose that the demand function for a product is given by $d(p)=\frac{60000}{p}$ and that the price $p$ is a function of time given by $p = 1.4t+12$ where $t$ is in days.\na) find the demand as a function of time $t$.\nb) find the rate of change of the quantity demanded when $t = 90$ days.\na) $d(t)=square$ (simplify your answer.)\nb) what is the approximate rate of change of the quantity demanded when $t = 90$ days?\n$square\frac{\text{units}}{\text{day}}$\n(simplify your answer. round to three decimal places as needed.)

suppose that the demand function for a product is given by $d(p)=\frac{60000}{p}$ and that the price $p$ is a function of time given by $p = 1.4t+12$ where $t$ is in days.\na) find the demand as a function of time $t$.\nb) find the rate of change of the quantity demanded when $t = 90$ days.\na) $d(t)=square$ (simplify your answer.)\nb) what is the approximate rate of change of the quantity demanded when $t = 90$ days?\n$square\frac{\text{units}}{\text{day}}$\n(simplify your answer. round to three decimal places as needed.)

Answer

Explanation:

Step1: Substitute price - time function into demand - price function

Given $D(p)=\frac{60000}{p}$ and $p = 1.4t+12$, substitute $p$ into $D(p)$: $D(t)=\frac{60000}{1.4t + 12}$

Step2: Differentiate $D(t)$ using the quotient rule

The quotient rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 60000$, $u^\prime=0$, $v = 1.4t + 12$, and $v^\prime=1.4$. $D^\prime(t)=\frac{0\times(1.4t + 12)-60000\times1.4}{(1.4t + 12)^{2}}=-\frac{84000}{(1.4t + 12)^{2}}$

Step3: Evaluate $D^\prime(t)$ at $t = 90$

Substitute $t = 90$ into $D^\prime(t)$: $D^\prime(90)=-\frac{84000}{(1.4\times90 + 12)^{2}}=-\frac{84000}{(126+12)^{2}}=-\frac{84000}{138^{2}}=-\frac{84000}{19044}\approx - 4.411$

Answer:

a) $D(t)=\frac{60000}{1.4t + 12}$ b) $-4.411$ units/day