the total time in minutes that the operators at a call center are on the phone with customers t days after the call center goes live is modeled by $f(t) = 5t^3 + 16t + 16$. the total number of customers calling t days after the call center goes live is modeled by $c(t) = 8t^2 + 5t + 5$. the average time spent on the phone per customer is $g(t) = \frac{f(t)}{c(t)}$.
how fast is the average time spent on the phone per customer changing on the tenth day?
(round to two decimal places.)
minutes per customer per day

Question

Accepted Answer

# Explanation:
To find how fast the average time spent on the phone per customer is changing, we need to find the derivative of $ g(t)=\frac{f(t)}{c(t)} $ using the quotient rule. The quotient rule states that if $ g(t)=\frac{u(t)}{v(t)} $, then $ g^\prime(t)=\frac{u^\prime(t)v(t)-u(t)v^\prime(t)}{[v(t)]^2} $.

### Step 1: Identify $ u(t) $, $ v(t) $, $ u^\prime(t) $, and $ v^\prime(t) $
- Let $ u(t) = f(t)=5t^{3}+16t + 16 $, so $ u^\prime(t)=15t^{2}+16 $ (using the power rule: $ \frac{d}{dt}(t^{n})=nt^{n - 1} $)
- Let $ v(t)=c(t)=8t^{2}+5t + 5 $, so $ v^\prime(t)=16t + 5 $ (using the power rule)

### Step 2: Apply the quotient rule
$$
g^\prime(t)=\frac{(15t^{2}+16)(8t^{2}+5t + 5)-(5t^{3}+16t + 16)(16t + 5)}{(8t^{2}+5t + 5)^{2}}
$$

### Step 3: Evaluate at $ t = 10 $
First, calculate the numerator:
- Calculate $ (15t^{2}+16)(8t^{2}+5t + 5) $ at $ t = 10 $:
  - $ 15(10)^{2}+16=1500 + 16=1516 $
  - $ 8(10)^{2}+5(10)+5=800+50 + 5=855 $
  - $ 1516\times855=1516\times(800 + 50+5)=1516\times800+1516\times50 + 1516\times5=1212800+75800+7580 = 1296180 $
- Calculate $ (5t^{3}+16t + 16)(16t + 5) $ at $ t = 10 $:
  - $ 5(10)^{3}+16(10)+16=5000+160 + 16=5176 $
  - $ 16(10)+5=165 $
  - $ 5176\times165=5176\times(160 + 5)=5176\times160+5176\times5=828160+25880 = 854040 $
- Subtract the two results: $ 1296180-854040 = 442140 $

Next, calculate the denominator:
- $ (8t^{2}+5t + 5)^{2} $ at $ t = 10 $:
  - $ 8(10)^{2}+5(10)+5=800 + 50+5=855 $
  - $ 855^{2}=731025 $

Now, $ g^\prime(10)=\frac{442140}{731025}\approx0.605 $ (rounded to three decimal places for intermediate step, then we will round to two decimal places as required)

Wait, let's recalculate the numerator more accurately:

First, expand $ (15t^{2}+16)(8t^{2}+5t + 5) $:
$$
\begin{align*}
&15t^{2}(8t^{2}+5t + 5)+16(8t^{2}+5t + 5)\
=&120t^{4}+75t^{3}+75t^{2}+128t^{2}+80t + 80\
=&120t^{4}+75t^{3}+203t^{2}+80t + 80
\end{align*}
$$

Expand $ (5t^{3}+16t + 16)(16t + 5) $:
$$
\begin{align*}
&5t^{3}(16t + 5)+16t(16t + 5)+16(16t + 5)\
=&80t^{4}+25t^{3}+256t^{2}+80t + 256t + 80\
=&80t^{4}+25t^{3}+256t^{2}+336t + 80
\end{align*}
$$

Now, subtract the two expanded polynomials:
$$
\begin{align*}
&(120t^{4}+75t^{3}+203t^{2}+80t + 80)-(80t^{4}+25t^{3}+256t^{2}+336t + 80)\
=&120t^{4}-80t^{4}+75t^{3}-25t^{3}+203t^{2}-256t^{2}+80t - 336t+80 - 80\
=&40t^{4}+50t^{3}-53t^{2}-256t
\end{align*}
$$

Now, evaluate the numerator at $ t = 10 $:
$$
\begin{align*}
&40(10)^{4}+50(10)^{3}-53(10)^{2}-256(10)\
=&40\times10000+50\times1000-53\times100 - 2560\
=&400000+50000 - 5300-2560\
=&450000-(5300 + 2560)\
=&450000 - 7860\
=&442140
\end{align*}
$$

Denominator at $ t = 10 $:
$$
(8(10)^{2}+5(10)+5)^{2}=(800 + 50+5)^{2}=855^{2}=731025
$$

So $ g^\prime(10)=\frac{442140}{731025}\approx0.605 \approx0.61 $ (rounded to two decimal places)

# Answer:
$ 0.61 $ minutes per customer per day

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QUESTION IMAGE

Question

Explanation:

Step 1: Identify \( u(t) \), \( v(t) \), \( u^\prime(t) \), and \( v^\prime(t) \)

Step 2: Apply the quotient rule

Step 3: Evaluate at \( t = 10 \)

Answer: