Question

the number of customers at a restaurant ( t ) hours after opening can be modeled by ( q(t) = -2.5t^2 + 10t - 8 ) and the average amount that each customer spends, in dollars, at time ( t ) can be modeled by ( p(t) = 2t + 50 ) where ( 0 leq t leq 8 ). find the rate of change of revenue when the restaurant has been open for 4 hours. the revenue is ? by ? after the restaurant is open for 4 hours.

Explanation:

Step1: Define Revenue Function

Revenue \( R(t) \) is the product of number of customers \( q(t) \) and average spending per customer \( p(t) \). So, \( R(t)=q(t)\times p(t)=(-2.5t^{2}+10t - 8)(2t + 50) \).

Step2: Find Derivative of Revenue

Use product rule: If \( R(t)=u(t)v(t) \), then \( R^\prime(t)=u^\prime(t)v(t)+u(t)v^\prime(t) \).

• \( u(t)=-2.5t^{2}+10t - 8 \), \( u^\prime(t)=-5t + 10 \)
• \( v(t)=2t + 50 \), \( v^\prime(t)=2 \)

So, \( R^\prime(t)=(-5t + 10)(2t + 50)+(-2.5t^{2}+10t - 8)(2) \)

Step3: Simplify Derivative

First, expand \( (-5t + 10)(2t + 50) \):
\( (-5t)(2t)+(-5t)(50)+10(2t)+10(50)=-10t^{2}-250t + 20t + 500=-10t^{2}-230t + 500 \)

Then, expand \( (-2.5t^{2}+10t - 8)(2) \):
\( -5t^{2}+20t - 16 \)

Now, add the two results:
\( R^\prime(t)=(-10t^{2}-230t + 500)+(-5t^{2}+20t - 16)=-15t^{2}-210t + 484 \)

Step4: Evaluate at \( t = 4 \)

Substitute \( t = 4 \) into \( R^\prime(t) \):
\( R^\prime(4)=-15(4)^{2}-210(4)+484=-15\times16-840 + 484=-240-840 + 484=-596 \) Wait, this seems wrong. Wait, maybe miscalculation. Let's re - calculate the derivative.

Wait, let's recalculate \( R(t)=(-2.5t^{2}+10t - 8)(2t + 50) \)

First, multiply term by term:

\( -2.5t^{2}\times2t=-5t^{3} \)

\( -2.5t^{2}\times50=-125t^{2} \)

\( 10t\times2t = 20t^{2} \)

\( 10t\times50 = 500t \)

\( -8\times2t=-16t \)

\( -8\times50=-400 \)

So, \( R(t)=-5t^{3}-125t^{2}+20t^{2}+500t-16t - 400=-5t^{3}-105t^{2}+484t - 400 \)

Now, find derivative \( R^\prime(t)=-15t^{2}-210t + 484 \)

Wait, when \( t = 4 \):

\( -15\times(4)^{2}=-15\times16=-240 \)

\( -210\times4=-840 \)

\( -240-840 + 484=-596 \). But this is a negative rate of change. Wait, maybe the original functions are misread. Let's check the original problem again.

Wait, the number of customers \( q(t)=-2.5t^{2}+10t - 8 \). Let's check \( q(4)=-2.5\times16 + 40-8=-40 + 40 - 8=-8 \). That can't be, number of customers can't be negative. Maybe there is a typo in the problem, but assuming the functions are correct.

Wait, maybe I made a mistake in the product rule. Let's re - do the product rule:

\( u=-2.5t^{2}+10t - 8 \), \( u^\prime=-5t + 10 \)

\( v = 2t+50 \), \( v^\prime = 2 \)

\( R^\prime=u^\prime v+uv^\prime=(-5t + 10)(2t + 50)+(-2.5t^{2}+10t - 8)\times2 \)

Expand \( (-5t + 10)(2t + 50) \):

\( -5t\times2t=-10t^{2} \)

\( -5t\times50=-250t \)

\( 10\times2t = 20t \)

\( 10\times50 = 500 \)

So, \( -10t^{2}-250t + 20t + 500=-10t^{2}-230t + 500 \)

Expand \( (-2.5t^{2}+10t - 8)\times2=-5t^{2}+20t - 16 \)

Now, add them: \( (-10t^{2}-230t + 500)+(-5t^{2}+20t - 16)=-15t^{2}-210t + 484 \)

Now, substitute \( t = 4 \):

\( -15\times16=-240 \)

\( -210\times4=-840 \)

\( -240-840 + 484=-596 \)

So the rate of change of revenue at \( t = 4 \) is - 596 dollars per hour.

Answer:

-596