Question

the number of customers at a restaurant ( t ) hours after opening can be modeled by ( q(t) = -3.5t^2 + 10t - 8 ) and the average amount that each customer spends, in dollars, at time ( t ) can be modeled by ( p(t) = 6t + 45 ) where ( 0 leq t leq 8 ). find the rate of change of revenue when the restaurant has been open for 6 hours. the revenue is decreasing by per hour after the restaurant is open for 6 hours.

Explanation:

Step1: Recall the revenue function

Revenue \( R(t) \) is the product of the number of customers \( q(t) \) and the average amount each customer spends \( p(t) \). So \( R(t)=q(t)\times p(t) \). Given \( q(t)= - 3.5t^{2}+10t - 8 \) and \( p(t)=6t + 45 \).

Step2: Find the derivative of \( R(t) \) using the product rule

The product rule states that if \( R(t)=u(t)v(t) \), then \( R^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t) \).
First, find \( q^\prime(t) \):
\( q(t)=-3.5t^{2}+10t - 8 \), so \( q^\prime(t)=\frac{d}{dt}(-3.5t^{2}+10t - 8)=-7t + 10 \)
Then, find \( p^\prime(t) \):
\( p(t)=6t + 45 \), so \( p^\prime(t)=\frac{d}{dt}(6t + 45)=6 \)

Now, apply the product rule:
\( R^\prime(t)=q^\prime(t)p(t)+q(t)p^\prime(t)=(-7t + 10)(6t + 45)+(-3.5t^{2}+10t - 8)\times6 \)

Step3: Evaluate \( R^\prime(t) \) at \( t = 6 \)

First, substitute \( t = 6 \) into \( q^\prime(t) \) and \( p(t) \):
\( q^\prime(6)=-7\times6 + 10=-42 + 10=-32 \)
\( p(6)=6\times6+45 = 36 + 45 = 81 \)
Substitute \( t = 6 \) into \( q(t) \) and \( p^\prime(t) \):
\( q(6)=-3.5\times6^{2}+10\times6 - 8=-3.5\times36 + 60 - 8=-126+60 - 8=-74 \)
\( p^\prime(6)=6 \)

Now, calculate \( R^\prime(6) \):
\( R^\prime(6)=q^\prime(6)p(6)+q(6)p^\prime(6)=(-32)\times81+(-74)\times6 \)
\(=-2592-444=-3036\)? Wait, that can't be right. Wait, maybe I made a mistake in the product rule application. Wait, no, wait the problem says "the rate of change of revenue". Wait, maybe I misread the problem. Wait, the number of customers is \( q(t)=-3.5t^{2}+10t - 8 \), and the average amount per customer is \( p(t)=6t + 45 \). So revenue \( R(t)=q(t)\times p(t) \). But maybe the question is about the rate of change of revenue with respect to time, so we need to find \( R^\prime(t) \) and then plug \( t = 6 \). But let's recalculate \( R^\prime(t) \) correctly.

First, expand \( (-7t + 10)(6t + 45) \):
\( (-7t)(6t)+(-7t)(45)+10\times6t + 10\times45=-42t^{2}-315t + 60t + 450=-42t^{2}-255t + 450 \)

Then, expand \( (-3.5t^{2}+10t - 8)\times6=-21t^{2}+60t - 48 \)

Now, add the two expansions:
\( R^\prime(t)=(-42t^{2}-255t + 450)+(-21t^{2}+60t - 48)=-63t^{2}-195t + 402 \)

Now, substitute \( t = 6 \) into \( R^\prime(t) \):
\( R^\prime(6)=-63\times6^{2}-195\times6 + 402=-63\times36-1170 + 402=-2268-1170 + 402=-3036 \)

Wait, that's a large negative number, but the problem says "the revenue is decreasing by...". Wait, maybe I made a mistake in the function definitions. Wait, the number of customers at time \( t \) is \( q(t)=-3.5t^{2}+10t - 8 \), and the average amount per customer is \( p(t)=6t + 45 \). Let's check the value of \( q(6) \): \( -3.5\times36 + 60 - 8=-126 + 60 - 8=-74 \). Wait, a negative number of customers? That doesn't make sense. Maybe there's a typo in the problem, but assuming the functions are correct, let's proceed.

Wait, maybe the question is to find the rate of change of revenue, which is \( R^\prime(t) \), and at \( t = 6 \), we have \( R^\prime(6)=-3036 \)? But that seems odd. Wait, maybe I misapplied the product rule. Let's try another approach. The rate of change of revenue is also equal to \( q^\prime(t)p(t)+q(t)p^\prime(t) \). Let's recalculate \( q^\prime(6) \): \( q^\prime(t)=-7t + 10 \), so \( q^\prime(6)=-42 + 10=-32 \). \( p(6)=6\times6 + 45=81 \). \( q(6)=-3.5\times36 + 60 - 8=-126 + 60 - 8=-74 \). \( p^\prime(t)=6 \). So \( R^\prime(6)=(-32)\times81+(-74)\times6=-2592 - 444=-3036 \). So the rate of change of revenue is -3036 dollars per hour, meaning the revenue is decreasing by 3036 dollars per hour? That seems very high, but maybe the numbers in the problem are…

Answer:

3036 (since it's decreasing by 3036 dollars per hour)